Answer:
B
Step-by-step explanation:
(8,12)=(x1, y1)
(1,3)=(X2, y2)
distance=√(x2-x1)^2+(y2-y1)^2
=√(1-8)^2+(3-12)^2
=√(49+81)
√130
=11.40
-5
What is the sign of a.
()
when a = 0 and b < 0?
b
Sos I need help
[tex]\qquad\qquad\huge\underline{\boxed{\sf Answer}}[/tex]
The sign of a will be negative ~
let the value of be be ' -x ' , -x is a negative number (b = -x < 0)
[tex]\qquad \sf \dashrightarrow \: a \times \frac{ - b}{b} [/tex]
[tex]\qquad \sf \dashrightarrow \: a \times \frac{ - ( - x)}{ - x} [/tex]
[tex]\qquad \sf \dashrightarrow \: a \times \frac{ x}{ - x} [/tex]
[tex]\qquad \sf \dashrightarrow \: a \times \frac{ - x}{ x} [/tex]
[tex]\qquad \sf \dashrightarrow \:a \times - 1[/tex]
[tex] \qquad \sf \dashrightarrow \:- a[/tex]
Therefore, the term a, has a negative sign unless it's value is substituted, I.e -a = 0 ( having no sign )
Suppose you roll a number cube. What is the probability that you roll the number 5?
Answer:
There is 1/6 chance that you will roll a 5
Step-by-step explanation:
Hope that helps!
Workers in an office of 60 staff were asked their favourite type of take-away.
The results are summarised in the table.
Pizza=8
Curry=8
Fish & chips=5
Kebab=9
Other=30
Work out the size of each angle to draw a pie chart.
PA ANSWERR NEED SOME ANSWERS THNKS
Answer:
16. j = 105 °
17. Q = 68 °
18. s = 97 °
19. t = 70 °
20. r = 125 °
Step-by-step explanation:
16. Name the angle that is supplementary to angle J as K, and label the angle that is next to K as G
Angle G = 105, because it is corresponding to the 105 angle.
Since G is supplementary to K, angle K = 75.
So, J = 105
17. Name the angle that is supplementary to angle Q as R
Angle R = 112, because it is corresponding to the 112 angle.
Since R is supplementary to Q, angle Q = 68.
18. Name the angle that is supplementary to the 83 angle as L
Angle L = 97, since angle L and the 83 angle are supplementary angles
-> Angle S = 97, since S and L are alternate angles
19. Angle T = 70, since angle T and the 70 angle are corresponding angles.
20. Name the angle that are alternate angles with angle R as Q
Angle Q = 125, because it is corresponding to the 125 angle.
Since angle Q and angle R are alternate angles, angle R = 125.
(Note : alternate angles and corresponding angles are equal to each other, supplementary angles add up to 180.)
Find the area of the figure
Answer:
C. 418 cm²
Step-by-step explanation:
Rectangle
= s × s
= 19 × 19
= 361 cm²
Triangle
= 1/2 × 6 × 19
= 57 cm²
361 + 57 = 418 cm²
So, therefore your answer is 418 cm²
Ella traveled by train to visit her friend.
• It took 3 hours for Ella to reach her destination.
• The train averaged a speed of 100 miles per hour.
Enter an equation to express the distance, d, Ella traveled.
Answer:
d = st, d = 100 x 3
Step-by-step explanation:
distance = speed x time
distance = 100 x 3, therefore, distance = 300, but if you just want the equation, it is d = 100 x 3 or d = st
Step-by-step explanation:
d = 100 x 3.
d= 100n
where n = time the journey took.
Calculate the area of the equilateral triangle using the formula for area of a regular polygon, and compare it to Bianca’s answer. The apothem, rounded to the nearest tenth, is units. The perimeter of the equilateral triangle is units. Therefore, the area of the equilateral triangle is , or approximately 43. 5 units2. The calculated areas are.
The calculated areas of both are the same instead of using the different formulas that are 43. 5 squared unit.
What is the area of a regular polygon?The area of the polygon is the product of half of the apothem of the triangle and perimeter.
The area of the polygon = [tex]\frac{1}{2} \times h \times p[/tex]
The side of the triangle is 10 units.
Apothem of an equilateral triangle
[tex]s = \frac{\sqrt{3} a}{6}[/tex]
s= [tex]\sqrt{3} \times 10/6[/tex]
s = 2.88
Here, a is the side of the triangle.
Thus the apothem of the triangle is 2.9 units.
The perimeter of the equilateral triangle is equal to the three times the sides.
Perimeter of the equilateral triangle ,
P = 3(10)
P = 30
Therefore, the perimeter of the equilateral triangle is 30 units.
The area of the equilateral triangle-
The area of the polygon = [tex]\frac{1}{2} \times h \times p[/tex]
A = [tex]1/2 \times 2.9 \times 30[/tex]
A = 43.5
Thus the area of the equilateral triangle is 43.5 squared units.
Therefore, The calculated areas of both are the same instead of using the different formulas.
Learn more about the apothem;
brainly.com/question/10580427
Answer:
(A) 2.9
(C)30
(A)1/2(2.9)(30)
(D)The same despite using different formulas
Step-by-step explanation:
I did it
i need help. please
Answer:
[tex]V=\pi r^2h=\pi *6^2*19[/tex] ≈ [tex]2148.8[/tex]
Let me know if you found this helpful! It means a lot!
A boy yells in a narrow valley. the sound reflects off of a boulder that is 80 m away. how long will it take to hear the echo if the speed of a sound wave is 320 m/s? (speed = ) 0.25 seconds 0.50 seconds 2.0 seconds 4.0 seconds
Answer:
0.50 seconds
Step-by-step explanation:
The time it takes for an echo to return is the time it takes for the sound to make the round trip between the source and the reflecting object.
The boulder is 80 m away from the source, so the round-trip distance is 2×80 m = 160 m. At a speed of 320 m/s, it takes sound ...
(160 m)/(320 m/s) = 0.50 s
to make the round-trip.
It will take 0.50 seconds to hear the echo.
The time it will take to hear the echo, when the speed of a sound wave is 320 m/s is 0.50 seconds.
What is the speed of sound?
Speed of sound is the rate by which the sound wave travels. The speed of sound depends on the medium by which the sound wave is travelling.
The speed of the wave can be found out using the following formula.
[tex]s=\dfrac{d}{t}[/tex]
Here, (d) is the distance travelled by the wave and (t) is time taken by the wave to cover that distance.
A boy yells in a narrow valley. The sound reflects off of a boulder that is 80 m away.
For the echo, distance travelled will be doubled as it come back.Thus, the distance travelled by sound wave is,
[tex]d=2\times80\\d=160\rm\; m[/tex]
The speed of sound is 320 m/s. Put these value in the above formula as,
[tex]320=\dfrac{160}{t}\\t=0.5\rm\; s[/tex]
Thus, the time it will take to hear the echo, when the speed of a sound wave is 320 m/s is 0.50 seconds.
Learn more about the speed of sound here;
https://brainly.com/question/95976
Evaluate the function for the given value of x.
Answer:
f(-6)= -41 923
Step-by-step explanation:
f(-6)= -( -6)⁶ +5(-6)⁴+8(-6)³+2(-6) -7 you substitute x with -6 the you use a calculator which will then give you - 41923
Guys, I’m giving you at least 100 points for you to answer to my angles, triangles, and quadrilaterals word problem homework package. So guys, please listen up and pay attention to me without ignoring my angles, triangles, and quadrilaterals word problem homework package for no reason! :D
Please read my 2 of the geometric word problem questions carefully in order to get the clear understanding in order for you to get the Brainliest award if your answer is the best efforts and step-by-step explanations all of the time and here’s an image for you to answer to my 2 of my geometric word problem questions.
Please answer to my geometric word problem questions as soon as possible with the best efforts and the best step-by-step explanations in order to get the Brainliest award of all the time! :D
Well anyways, guys, good luck on answering 3 math word problem questions of my my angles, triangles, and quadrilaterals word problem homework package in order to get the best Brainliest award of all of the time, and I will check the answers to see if it’s appropriate for me. :)
Step-by-step explanation:
For question 22: Notice the congruent symbol on sides AB and CB; this shows that the triangle is isosceles, and in an isosceles triangle, the angles opposite congruent sides are congruent. Thus, angle A actually equals angle C.
In other the find B, it is know that all of the interior angles of a triangle must add to 180; thus, this can be written as: angle A + angle B + angle C = 180.
Since you already know angles A and C, subtract angle A and angle C from 180 to get angle B.
For question 23:
Notice that triangle DBC is an isosceles triangle, and that angle D and angle C are opposite of congruent sides, indicating that they are equal. Since the interior angles of a triangle must add up to 180, and you know that angle D equals angle C, you can find angle DBC by doing 180 - (2 * angle D).
Now that you have angle DBC, you can find angle CBA; since these two angles form a straight line, you know they must add to 180 degrees. Thus, 180 - angle DBC equals angle CBA.
Now, notice that triangle CBA is also isosceles, and that angles CBA and angle A are opposite congruent sides, indicating that they are also congruent. Thus, angle A equals angle CBA.
Hope this helps :)
pls can I get some help this is compound interest (10) will give brainliest
Answer:
2.7
Step-by-step explanation:
I need help ASAP...
(reward 10 points)
Answer:
it may be C im sorry if its wrong
Step-by-step explanation:
please mark me the brainiest
Answer:
The answer is a negative correlation.
Step-by-step explanation:
It is a negative correlation because when the y variable tends to decrease as the x variable increases, you can say there is a negative correlation between the variables.
quadrilateral ABCD is a kita with diagonals AC and BD. determine the area of a kita with diagonals BD = 8 units and AC = 12 units
[tex]\mathsf\blue{♧ANSWER♧}[/tex]
[tex] \mathsf \orange{ \frac{x \times y}{2}} [/tex]
[tex] \mathsf \green{ \frac{8 \times 12}{2} }[/tex]
[tex] \mathsf \orange{ \frac{96}{2} }[/tex]
[tex] \mathsf \green{48 \: square \: units}[/tex]
[tex] \huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }[/tex]
[tex]\qquad\qquad\qquad \large\underline{ \boxed{ \sf{✰\: Information }}}[/tex]
★ Given➣A quadrilateral ABCD shape (kite) with diagonals "AC" and "BD"➣ diagonal BD is of 8units ➣ diagonal AC is of 12units✜ Now let's see about ques ✜question is asking about to find area of kite with the above infosoo we know that[tex]{ \boxed{ ✟ \: \underline{ \boxed{ \sf \: Area \: of \: kite = \frac{ d_1 \times d_2}{2} \: or \blue{ \: \frac{1}{2} \times d_1 \times d_2}}}✟ }}[/tex]
[tex]\rule{80mm}{2.5pt}[/tex]
✜ Let's substitute value according to formula ✜[tex] \qquad \rm{➛Area \: of \: kite = \frac{BD×AC}{2} }[/tex]
[tex]\qquad \rm{➛Area \: of \: kite = \frac{8×12}{2} } \\ \qquad \rm{➛Area \: of \: kite = \frac{ \cancel{96}}{ \cancel2} } \\ \qquad \rm{➛Area \: of \: kite = 48 {units}^{2} }[/tex]
[tex]\rule{80mm}{2.5pt}[/tex]
★Hence area of kite with diagonals BD = 8 units and AC = 12 units is
[tex]\qquad\qquad\qquad{ \boxed{↪ \underline{ \boxed{ \sf{\: 48 {units}^{2} \green✓ }}}↩}}[/tex]
[tex]\rule{80mm}{2.5pt}[/tex]
★ More info regarding this topic ★
➣ Quadrilateral:- A polygon with four sides,sum of it's interior angle is of 360⁰➣ Kite is also a quadrilateral because of four sides.Kite is a quadrilateral which is proved Euclidean geometry.[tex]\rule{80mm}{2.5pt}[/tex]
Hope it helps !
Find the value of cos⁻¹ (cos7π/6).
Answer:
5π/6
Step-by-step explanation:
Given :
cos⁻¹ (cos7π/6)Solving :
We know that : cos⁻¹ cosθ = θBut we can't just do that in this caseBecause the range of cos values is [0, π]Clearly, our value does not lie in this rangeWe have to take a different Quadrant other than the 3rd Quadrant which gives cos a negative valueThe 2nd Quadrant also has cos values negativeTherefore,cos⁻¹ cos (π - π/6)cos⁻¹ cos (5π/6)⇒ 5π/6 ∈ [0, π] ⇒ It lies in the range!Answer:
5π Over 6
Step-by-step explanation:
You can write 7π6 as (π+π6)Thus we can clearly see the angle falls in the third quadrant. And the cosine value in third quadrant is always negative. Hence, cos(π+π6)=−cos(π6)coming back to the question cos−1[cos(7π6)]=cos−1[−cos(π6)]=π−cos−1[cos(π6)]=π−π6=5π6
one trillion one hundred eighty-six billion two hundred million in number form
Step-by-step explanation:
1,186,200,000,000 in standard form
Help please thank youuu
Answer:
18 mi^2
...
...
...
...
...
Answer:
42 square miles
Step-by-step explanation:
The area of a composite shape can be found by decomposing it into shapes whose area formulas you know. Here, the shape can be decomposed into a triangle and two rectangles. The two rectangles can be formed several ways.
TriangleThe area of the triangle at the top of the figure is ...
A = 1/2bh
A = 1/2(6 mi)(3 mi) = 9 mi²
Overall rectangleThe area of the overall rectangle (including the space at lower right) is ...
A = LW
A = (9 mi)(5 mi) = 45 mi²
Deleted spaceThe area of the rectangle at lower right that is not included in the shaded area is ...
A = LW
A = (4 mi)(3 mi) = 12 mi²
Total areaThe total area of the figure is the sum of the triangle and overall rectangle, less the excluded area at lower right.
total area = 9 mi² +45 mi² -12 mi² = 42 mi²
The area of the figure is 42 square miles.
_____
Additional comment
Other ways to divide the figure into two rectangles are ...
extend the vertical line to make a 9×2 and a 3×5 rectangleextend the horizontal line to make a 4×2 and a 5×5 rectangleThe (period,amplitude,frequence) of y=1/4cos1/2x is 4pie.
one of these is the answer need help
Answer:
period
Step-by-step explanation:
The attached graph shows a period of the function. Its amplitude is 1/4 (vertical distance from centerline to peak). Its period is 4π (horizontal distance from peak to peak). Its frequency is the reciprocal of the period, so would be 1/(4π).
The value 4π is the period of the function.
Which is a perfect square?
O 5
O 8
O 36
044
Answer:
6×6=36
36 is the perfect square
m exponent 2 plus 15 if m =4
Answer:
31
Step-by-step explanation:
[tex]4^{2}[/tex]=16
16+15=31
CHOOSE THE CORRECT ANSWER
PLEASE HELP!
Solve for x. 4/5 = x/7 x =
Answer:
5.6
Step-by-step explanation:
4/5 = .8
so ? ÷ 7 = .8|
7 x .8 = 5.6
x= 5.6
x = 5.6
hope it helps...!!!!
Find the value of a. Round your answer to the nearest tenth. B 10 77.5 A C 7.9
Answer:
[tex]\displaystyle 11,3[/tex]
Step-by-step explanation:
Use the Law of Cosines to find the length of the third edge:
Solving for Angles
[tex]\displaystyle \frac{a^2 + b^2 - c^2}{2ab} = cos\angle{C} \\ \frac{a^2 - b^2 + c^2}{2ac} = cos\angle{B} \\ \frac{-a^2 + b^2 + c^2}{2bc} = cos\angle{A}[/tex]
Use [tex]\displaystyle cos^{-1}[/tex]towards the end or you will throw your result off!
Solving for Edges
[tex]\displaystyle b^2 + a^2 - 2ba\:cos\angle{C} = c^2 \\ c^2 + a^2 - 2ca\:cos\angle{B} = b^2 \\ c^2 + b^2 - 2cb\:cos\angle{A} = a^2[/tex]
Take the square root of the final result or it will be thrown off!
Let us get to wourk:
[tex]\displaystyle 10^2 + 7,9^2 - 2[10][7,9]\:cos\:77,5 = a^2 \hookrightarrow 100 + 62,41 - 158\:cos\:77,5 = a^2 \hookrightarrow \sqrt{128,212541} = \sqrt{a^2}; 11,323097677... \\ \\ \boxed{11,3 \approx a}[/tex]
I am joyous to assist you at any time.
help please i don’t know how to do it
Answer:
x + 3y = 15
Step-by-step explanation:
Line is passing through the points [tex](3, \:4) = (x_1,\:y_1)\:and\: (9, 2)=(x_2,\:y_2)[/tex]
Slope of line (m)= (2 - 4)/(9 - 3) = -2/6 = -1/3
Equation of line in point-slope form is given as:
[tex]y-y_1 =m(x-x_1)[/tex]
Plugging the values of [tex]x_1,\: y_1\: and \: m[/tex] in the above equation we find:
[tex]y-4 =-\frac{1}{3}(x-3)[/tex]
[tex]\implies 3y-12 =-x+3[/tex]
[tex]\implies x+ 3y =12+3[/tex]
[tex]\implies \huge {\orange{\boxed{x+ 3y = 15}}}[/tex]
This is the required equation of line in the form [tex]\red{\bold{ax + by = c}} [/tex]
Directions: Calculate the area of a circle using 3.14x the radius
1) d = 4.4 mm.
Calculate the area of the circle.
2) d = 3.7 cm.
Calculate the area of the circle.
3) r= 8.3 cm.
Calculate the area of the circle.
4) d = 5.8 yd.
Calculate the area of the circle.
5) d = 1 yd.
Calculate the area of the circle
6) r = 8 ft.
Calculate the area of the circle
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
As we know ~
Area of the circle is :
[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
And radius (r) = diameter (d) ÷ 2
[ radius of the circle = half the measure of diameter ]
➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖
Problem 1[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 4.4\div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 2.2 \: mm[/tex]
Now find the Area ~
[tex]\qquad \sf \dashrightarrow \: \pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{} [/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
problem 2[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 3.7 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm[/tex]
Bow, calculate the Area ~
[tex]\qquad \sf \dashrightarrow \: \pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{} [/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
Problem 3[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 68.89[/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
Problem 4[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 5.8 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd[/tex]
now, let's calculate area ~
[tex]\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 8.41 [/tex]
[tex] \qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
problem 5[tex]\qquad \sf \dashrightarrow \:r = d \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 1 \div 2[/tex]
[tex]\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd[/tex]
Now, let's calculate area ~
[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 0.25[/tex]
[tex]\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2} [/tex]
・ ．━━━━━━━†━━━━━━━━━．・
problem 6[tex]\qquad \sf \dashrightarrow \:\pi {r}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2} [/tex]
[tex]\qquad \sf \dashrightarrow \:3.14 \times 64[/tex]
[tex]\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2} [/tex]
➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖
solve for x and y 13x−28=24
x=4
also uhh where does the y go-
Pls help me on this problem I’m confused
5 quarts I think 4 cups per quart has 5 quarts and 2 cups I think
Answer:
1 quart = 4 cups
8 × 4 = 32
32 - 6 = 26
26 - 18 = 8
he has 2 quarts left
please help asap have a good day and ily
Answer:
Pretty sure it's 2:30 P.M.
Step-by-step explanation:
If her track practice started at 12:00 P.M., and and ran relays for one hour, it would already be 1:00 P.M.
Then she practiced the long jump for an hour and thirty minutes. Add another hour to that and it would be 2:00 P.M. Add thirty minutes, and it would be 2:30 P.M.
Hopefully that makes sense to you
Lines a and b are parallel.
what is the value of x?
A) 15
B) 16
C) 56
D) 64
Answer:
56
Step-by-step explanation:
What is the value of x? (Do not put in the degree symbol)
Step-by-step explanation:
x+134=180(straight angle)
x=46