Indicate in standard form the equation of the line passing through the given points, writing the answer in the equation box below.

G(4, 6), H(1, 5)

Answer:

3 + 2k

Step-by-step explanation:

Given:

1/5(15 + 10k)

= (1/5 * 15) + (1/5 * 10k)

= (1 * 15)/5 + (1 * 10k)/5

= 15/5 + 10k/5

= 3 + 2k

Therefore,

1/5(15 + 10k) = 3 + 2k

The equivalent expression is 3 + 2k

Answer:

a^2 -25

Step-by-step explanation:

5a-25 + a^2 - 5a

a^2 - 25

Answer:

C

Step-by-step explanation:

Answer:

2 units

Step-by-step explanation:

Using the given formula

P = 2(l + w)

  = 2([tex]\frac{2}{3}[/tex] + [tex]\frac{1}{3}[/tex] )

   = 2(1)

   = 2 units

Answer:

8.5

Step-by-step explanation:

QR is 10

PQRS AREA is 45

10 times 4.5 is 45

meaning BC and PQ is 4.5

4.5 plus 4.5 is 9

26 minus 9 is 17

17 divided by two is 8.5

AB is 8.5

Answer:

x ≈ 28.8°

Step-by-step explanation:

Using the sine ratio in the right triangle

sin x = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{AB}{AC}[/tex] = [tex]\frac{3.9}{8.1}[/tex] , then

x = [tex]sin^{-1}[/tex] ([tex]\frac{3.9}{8.1}[/tex] ) ≈ 28.8° ( to the nearest tenth )

Answer:

A

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Total waste = 22 000 in^3

If we use a proportion we have:

x : 100 = 2000 : 22 000

x = 200 000 / 22 000 = 9,09 % (option a is wrong

x : 100 = 7000 : 22 000

x = 700 000 / 22 000 = 31 % (ok)

Answer:

10cm2

Step-by-step explanation:

A=1÷2×4×5=10cm2

Answer:

x = 50.6

Step-by-step explanation:

Answer:

Step-by-step explanation:

Picture this circumstance occurring within a circle. The center of the circle is where all the blades meet. That means that in terms of a circle's parts, these blades are the radii of the circle. And if they are all evenly spaced out, and there are 3 of them, the central angle in between each pair is 120 degrees (360/3). Once we know that, we can use the arc length formula for a circle which is

[tex]AL=\frac{\theta}{360}*2\pi r[/tex] and filling in:

[tex]AL=\frac{120}{360}*2(3.1415)(131)[/tex] so

AL = 274.4 feet

Answer:

12

Step-by-step explanation:

45-33 is 12

And I guess to check, make sure 12 < 18

Answer:

Opposite sides of a parallelogram are parallel (by definition) and so will never intersect. The area of a parallelogram is twice the area of a triangle created by one of its diagonals. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides.

Step-by-step explanation:

Answer:

(-12,2)

Step-by-step explanation:

x^2 + 10x = 24

x^2 + 10x + (10/2)^2 = 24 + (10/2)^2

10/2 = 5

5^2 = 25

x^2 + 10x + 25 = 24 + 25

x^2 + 10x + 25 = 49

(x + 5)^2 = 49                 Take the square root of both sides

(x + 5) = sqrt(49)

x + 5 = +/- 7

x = +/- 7 - 5

x = +7 - 5 = 2

x = -7 - 5 = -12

Answer:

{ -12 , 2}

Step-by-step explanation:

x² + 10x = 24

In order to complete the square, the equation must first be in the form x² + bx =c.

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

expand exponents.

Add 24 and 25

Factor x² + 10x + 25. In general, when x² + bx + c is a perfect square, it can always be factored as ( x + b/2)².

Take the square root of both sides of the equation.

[tex] \small \sf \sqrt{(x + 5) {}^{2} } = \sqrt{49} [/tex]

simplify

Subtract 5 from both sides.

x + 5 - 5 = 7 - 5

x + 5 - 5 = +/- 7 -5

Answer:

Angle E is the included angle between sides DE and EF.

The included angle between sides FE and DF is angle F

Step-by-step explanation:

Given

[tex]\triangle D EF[/tex]

Required

Complete the blanks

To solve this, I will attach an illustration of [tex]\triangle D EF[/tex]

From the attached image of [tex]\triangle D EF[/tex], we can see that"

[tex]\angle E[/tex] is between DE and EF

[tex]\angle F[/tex] is between FE and DF

Answer:

see picture

Step-by-step explanation:

Answer:

da

Step-by-step explanation:

Answer:

just be smart

Step-by-step explanation: (2x)

Answer:  43.1 degrees (choice B)

==================================================

Work Shown:

Use the law of sines

sin(A)/a = sin(C)/c

sin(A)/20 = sin(82)/29

sin(A) = 20*sin(82)/29

sin(A) = 0.68294349

A = arcsin(0.68294349)

A = 43.074088

A = 43.1 degrees

Answer:

43.1°

Step-by-step explanation:

Hello, just finished the quiz and the correct answer is 43.1 degrees.

Answer:

f = m+h

Step-by-step explanation:

just add h to both sides of the equation to isolate f.

Hope this helps!

Answer:

CF = 8.5

Step-by-step explanation:

Given 2 intersecting chords, then

The product of the parts of one chord is equal to the product of the parts of the other chord, that is

CF × DF = BF × EF , substitute values

CF × 14 = 7 × 17

14CF = 119 ( divide both sides by 14 )

CF = 8.5

This is because the opposite angles of any inscribed quadrilateral (aka cyclic quadrilateral) are always supplementary. So x+130 = 180 solves to x = 50.

If OP is circumscribed about quadrilateral ABCD then the value of x is 50 degrees, Option B is correct.

A quadrilateral is a four-sided polygon, having four edges and four corners.

A quadrilateral is inscribed insider a circle with centre p.

ABCD is a quadrilateral

We have to find the value of x.

As we know that the sum of the opposite side measures is 180 degrees.

m∠A + m∠C = 180

x + 130 =180

Subtract 130 from both sides

x = 180-130

x= 50 degrees.

Hence, if OP is circumscribed about quadrilateral ABCD then the value of x is 50 degrees, Option B is correct.

To learn more on Quadrilateral click:

https://brainly.com/question/29934440

#SPJ7

Given:

The expression is:

[tex]b^2+20b[/tex]

To find:

The a monomial so that the trinomial may be represented by a square of a binomial.

Solution:

If an expression is [tex]x^2+bx[/tex], then be need to add square of half of coefficient of x, i.e., [tex]\left(\dfrac{b}{2}\right)^2[/tex] in the given expression to make in perfect square.

We have,

[tex]b^2+20b[/tex]

Here, coefficient of b is 20,so wee need to add square of half of coefficient of b, i.e., [tex]\left(\dfrac{20}{2}\right)^2[/tex].

[tex]\left(\dfrac{20}{2}\right)^2=10^2[/tex]

[tex]\left(\dfrac{20}{2}\right)^2=100[/tex]

Therefore, we need to add 100 to make [tex]b^2+20b[/tex] a perfect square binomial.

Answer:

7/50 = x/75

x = 10.5

therefore he would run 10.5 miles in 75 minutes

Answer:

In 75 minutes, she runs (7/50) x 75 = 10.5.

She runs 10.5 miles in 75 minutes.

Step-by-step explanation:

Answer:

49

Step-by-step explanation:

Answer:

[tex]10[/tex].

Step-by-step explanation:

See below for a proof of why all but the first digit of this [tex]N[/tex] must be "[tex]9[/tex]".

Taking that lemma as a fact, assume that there are [tex]x[/tex] digits in [tex]N[/tex] after the first digit, [tex]\text{A}[/tex]:

[tex]N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}[/tex], where [tex]x[/tex] is a positive integer.

Sum of these digits:

[tex]\text{A} + 9\, x= 2021[/tex].

Since [tex]\text{A}[/tex] is a digit, it must be an integer between [tex]0[/tex] and [tex]9[/tex]. The only possible value that would ensure [tex]\text{A} + 9\, x= 2021[/tex] is [tex]\text{A} = 5[/tex] and [tex]x = 224[/tex].

Therefore:

[tex]N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}[/tex].

[tex]N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}[/tex].

[tex]N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}[/tex].

Hence, the sum of the digits of [tex](N + 2021)[/tex] would be [tex]6 + 2 + 2 = 10[/tex].

Lemma: all digits of this [tex]N[/tex] other than the first digit must be "[tex]9[/tex]".

Proof:

The question assumes that [tex]N\![/tex] is the smallest positive integer whose sum of digits is [tex]2021[/tex]. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of [tex]N[/tex] is not "[tex]9[/tex]".

For example: [tex]N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}[/tex], where [tex]\text{A}[/tex], [tex]\text{P}[/tex], [tex]\text{B}[/tex], and [tex]\text{C}[/tex] are digits. (It is easy to show that [tex]N[/tex] contains at least [tex]5[/tex] digits.) Assume that [tex]\text{B} \![/tex] is one of the non-leading non-"[tex]9[/tex]" digits.

Either of the following must be true:

If [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex], is a "[tex]0[/tex]", then let [tex]N^{\prime}[/tex] be [tex]N[/tex] with that "[tex]0\![/tex]" digit deleted: [tex]N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}[/tex].

The digits of [tex]N^{\prime}[/tex] would still add up to [tex]2021[/tex]:

[tex]\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}[/tex].

However, with one fewer digit, [tex]N^{\prime} < N[/tex]. This observation would contradict the assumption that [tex]N\![/tex] is the smallest positive integer whose digits add up to [tex]2021\![/tex].

On the other hand, if [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex], is not "[tex]0[/tex]", then [tex](\text{P} - 1)[/tex] would still be a digit.

Since [tex]\text{B}[/tex] is not the digit [tex]9[/tex], [tex](\text{B} + 1)[/tex] would also be a digit.

let [tex]N^{\prime}[/tex] be [tex]N[/tex] with digit [tex]\text{P}[/tex] replaced with [tex](\text{P} - 1)[/tex], and [tex]\text{B}[/tex] replaced with [tex](\text{B} + 1)[/tex]: [tex]N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}[/tex].

The digits of [tex]N^{\prime}[/tex] would still add up to [tex]2021[/tex]:

[tex]\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}[/tex].

However, with a smaller digit in place of [tex]\text{P}[/tex], [tex]N^{\prime} < N[/tex]. This observation would also contradict the assumption that [tex]N\![/tex] is the smallest positive integer whose digits add up to [tex]2021\![/tex].

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this [tex]N[/tex] other than the first digit must be "[tex]9[/tex]".

Therefore, [tex]N[/tex] would be in the form: [tex]N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}[/tex], where [tex]\text{A}[/tex], the leading digit, could also be [tex]9[/tex].

Answer:

I think its is C

Answer: 6

Step-by-step explanation: 6 is a median

Hey there!

To find your mean you have to put the numbers from descending (least/decreasing) to ascending (greater/increasing) order

Median is also understood as the “middle number”

2, 8, 6, 10, 4, 12​

= 2, 4, 6, 8, 10, 12

Based on that information, you have in your new data set you have 2 pairs in the middle (6 & 8), so you’ll have to total/average of the number. We have to DIVIDE them by 2 because it’s 2 numbers

6 ends the descending set and 8 starts the ascending set

Median = 6 + 8 / 2

6 + 8 = 14

= 14/2

= 7

Answer: therefore your MEDIAN is most likely 7

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

Answer:

D

Step-by-step explanation:

because there is an even number of negatives\

Hope this help have a good day

Answer:

4 th option

Step-by-step explanation:

The product of an even / odd amount of positive numbers is positive

The product of an even amount of negative numbers is positive.

The product of an odd amount of negative numbers is negative.

Option 1

The product of 1 positive and 3 negative numbers will be negative

Option 2

Similar to option 1

Option 3

The product of 3 positive and 1 negative will be negative

Option 4

The product of 2 negative and 2 positive numbers will be positive

Answer:

angles Q and Q' are congruent, angles R and R' are congruent

Step-by-step explanation:

For two triangles to be congruent, we need to show is two of the  

1. corresponding angles are congruent,

or,

2. all three pairs of corresponding sides are proportional,  

or,

3. two pairs of corresponding sides are proportional AND the included angles are congruent.  

And, angles are always congruent when a figure is dilated

(satisfies the first criterion that at least two corresponding angles are congruent)