Play the four digit 3,5,7,and 9 into the boxes pure in the position that would give the greatest results in the true numbers are multiplied

- 73X95
- 79X53
-97X35
-93X75

Answer:

[tex]Correct answer\\y=\frac{1}{2} x-4[/tex]

Step-by-step explanation:

[tex]y-\frac{-5+3}{2} =\frac{-1}{\frac{3-(-5)}{4-8} } (x-\frac{4+8}{2} )\\y+1=\frac{1}{2} (x-6)\\y=\frac{1}{2} x-4[/tex]

Answer:

x = 4

Step-by-step explanation:

Corresponding angles are congruent

4x + 44 = 6x + 36

44 = 2x + 36

8 = 2x

x = 4

Answer:

Step-by-step explanation:

20000000000+3000000000+900000000+60000+8000+400+7

Answer:

1190 m^3

Step-by-step explanation:

l = 1700 cm = 17 m

b = 14 m

h = 1000 cm = 10 m

Total volume = l × b × h

= 17 × 14× 10

= 2380 m^3

since it is half filled ,

Volume is half , so,

volume of water in pond = 2380 ÷ 2

= 1190 m^3

Answer:

Kayleigh will have a total of $4608.

Step-by-step explanation:

First, you use the formula, I=PRT (Interest=Principal, Rate, Time), then you distribute the numbers: (I=4500x0.8%x3) when you multiply them all, you get $108, then you lastly add 108 to 4500, and you get your final answer of $4608.

Kayleigh will earn $108 in interest over a period of 3 years.

Interest is the additional amount of money that is charged or earned on a principal amount of money. It is typically expressed as a percentage of the principal and is either charged when borrowing money or earned when investing or saving money.

To calculate the interest Kayleigh will earn in 3 years, we can use the formula for simple interest:

Interest = Principal x  Rate x Time

Given:

Principal = $4500

Rate = 0.8% = 0.008 (decimal form)

Time = 3 years

Plugging the values into the formula, we have:

Interest = $4500 x 0.008 x 3

Calculating the expression, we find:

Interest = $108

Therefore, Kayleigh will earn $108 in interest over a period of 3 years.

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Answer:

Exact Form: [tex]\frac{17}{6}[/tex]

Decimal Form: 2.83

Mixed Number Form: [tex]2 and 5/6[/tex]

Step-by-step explanation:

Answer:

[tex]1\frac{1}{10}[/tex]

Step-by-step explanation:

[tex]4\frac{1}{3}-2\frac{1}{2}[/tex]

[tex]4\frac{2}{6} -2\frac{3}{6}=1\frac{5}{6}[/tex]

[tex]1\frac{5}{6} =\frac{11}{6}[/tex]

[tex]1\frac{2}{3}=\frac{5}{3}[/tex]

[tex]\frac{11}{6}[/tex] ÷ [tex]\frac{5}{3} =[/tex]

[tex]\frac{11}{6} *\frac{3}{5}=[/tex]

[tex]\frac{11}{2}*\frac{1}{5} =[/tex]

[tex]=\frac{11}{10}[/tex]

[tex]=1\frac{1}{10}[/tex]

Hope this is helpful

Answer:

its true trust biggie on this one fam

Step-by-step explanation:

its true broski

Answer:

D. f(t) = 192,345(1.04)^t

Step-by-step explanation:

I took the test and it was right.

Also that is the original price and when you look at exponential functions, the starting point or original price is always first the then rate of increase. The table just shows how it increased in year 1 and 2.

Hope this helps. :)

A function assigns the values. The function that best represents the value of the real estate after t years is f(t) = 192,345(1.04)^t. Thus, the correct option is D.

A function assigns the value of each element of one set to the other specific element of another set.

The initial cost of industrial real estate is $192,345, while the cost after one year is $200,038.80. Therefore, the rate of appreciation is,

[tex]\$200,038.80 = \$192,345(1+R)^t\\\\\$200,038.80 = \$192,345(1+R)^1\\\\\dfrac{\$200,038.80}{\$192,345}=(1+R)^t\\\\1.04 = 1 + R\\\\R = 0.04[/tex]

Hence, the function that best represents the value of the real estate after t years is f(t) = 192,345(1.04)^t. Thus, the correct option is D.

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Answer:

[tex]\rm\displaystyle \theta = \frac{\pi}{4} [/tex]

Step-by-step explanation:

we want to find the acute angle θ (as θ is between (0,π/2)) formed by the line y=-2x+4 and y=3x-3 to do so we can consider the following formula:

[tex] \displaystyle\tan( \theta) = \bigg| \frac{ m_{2} - m_{1} }{1 + m_{1} m_{2} } \bigg | [/tex]

[tex] \rm \displaystyle \implies\theta = \arctan \left( \bigg | \frac{ m_{2} - m_{1} }{1 + m_{1} m_{2} } \bigg | \right)[/tex]

From the first equation we obtain that [tex]m_1[/tex] is -2 and from the second that [tex]m_2[/tex] is 3 therefore substitute:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 3 - ( - 2) }{1 + ( - 2) (3)} \bigg | \right)[/tex]

simplify multiplication:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 3 - ( - 2) }{1 + ( - 6)} \bigg | \right)[/tex]

simplify Parentheses:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 3 + 2 }{1 + ( - 6)} \bigg | \right)[/tex]

simplify addition:

[tex] \rm\displaystyle \theta = \arctan \left( \bigg | \frac{ 5 }{ - 5} \bigg | \right)[/tex]

simplify division:

[tex] \rm\displaystyle \theta = \arctan \left( | - 1| \right)[/tex]

calculate the absolute of -1:

[tex]\rm\displaystyle \theta = \arctan \left( 1\right)[/tex]

calculate the inverse function:

[tex]\rm\displaystyle \theta = \frac{\pi}{4} [/tex]

hence,

the angle θ which is formed by the line y = -2x+4 and y = 3x-3 is π/4

(for more info about the formula refer the attachment thank you!)

Consider two vector-valued functions,

[tex]\vec r(t) = \left\langle t, -2t+4\right\rangle \text{ and } \vec s(t) = \left\langle t, 3t-3\right\rangle[/tex]

Differentiate both to get the corresponding tangent/direction vectors:

[tex]\dfrac{\mathrm d\vec r(t)}{\mathrm dt} = \left\langle1,-2\right\rangle \text{ and } \dfrac{\mathrm d\vec s(t)}{\mathrm dt} = \left\langle1,3\right\rangle[/tex]

Recall the dot product identity: for two vectors [tex]\vec a[/tex] and [tex]\vec b[/tex], we have

[tex]\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)[/tex]

where [tex]\theta[/tex] is the angle between them.

We have

[tex]\langle1,-2\rangle \cdot \langle1,3\rangle = \|\langle1,-2\rangle\| \|\langle1,3\rangle\| \cos(\theta) \\\\ 1\times1 + (-2)\times3 = \sqrt{1^2 + (-2)^2} \times \sqrt{1^2+3^2} \cos(\theta) \\\\ \cos(\theta) = \dfrac{-5}{\sqrt5\times\sqrt{10}} = -\dfrac1{\sqrt2} \\\\ \implies \theta = \cos^{-1}\left(-\dfrac1{\sqrt2}\right) = \dfrac{3\pi}4[/tex]

Then the acute angle between the lines is π/4.

Answer:

[tex]MK\approx 87.7[/tex]

Step-by-step explanation:

By definition, similar polygons have corresponding sides in a constant proportion. Therefore, we can set up the following proportion (ratio of corresponding sides) to solve for [tex]MK[/tex]:

[tex]\frac{20}{13}=\frac{MK}{57},\\MK=\frac{57\cdot 20}{13}\approx \boxed{87.7}[/tex]

Answer:

87.7

Step-by-step explanation:

Like stated previously similar triangle have side lengths with common ratios

*Create a proportionality to solve for MK*

57/13 = MK / 20

Now solve for MK

Multiply each side by 20

57/13 * 20 = 87.7 ( rounded )

MK/20 * 20 = MK

We're left with MK = 87.7

Answer:

√20

Or

2√5

Step-by-step explanation:

It is very easy just substitute

It will be:

Answer:

ok so we can simpley this is 1:106

so we multiply

20*106=2120 this is cm so we make it meters...

21.2meters!!!

Hope This Helps!!!

Answer:

first

2/5m + 8/5

second

22/5

third

11

Answer:

A

Step-by-step explanation:

Answer:

A and B, every even number is divisible by two, C is odd

Answer:

4/3

Step-by-step explanation:

The formula is (y1-y2) /(x1-x2)

Answer:

slope = [tex]\frac{4}{3}[/tex]

Step-by-step explanation:

Calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (2, 2) and (x₂, y₂ ) = (- 1, - 2)

m = [tex]\frac{-2-2}{-1-2}[/tex] = [tex]\frac{-4}{-3}[/tex] = [tex]\frac{4}{3}[/tex]

2 Answers:

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

Explanation:

Let's go through the answer choices

---------------------------------

A)

The lowest point is y = 1, and the highest point is y = 27. Computing the midpoint gets us (yMin+yMax)/2 = (1+27)/2 = 28/2 = 14.

The midline of [tex]h(\theta)[/tex] is y = 14.

Consider that the parent function [tex]\cos(\theta)[/tex] has a midline of y = 0. The jump from y = 0 to y = 14 must mean [tex]h(\theta)[/tex] has a vertical shift of 14.

Choice A is true

---------------------------------

B)

The period is the length of the cycle. For the function [tex]h(\theta)[/tex], the function starts at [tex]\theta = 0[/tex] and repeats when it gets to [tex]\theta = \frac{\pi}{2}[/tex]. This is a difference of pi/2 units which is the period.

Choice B is false. It contradicts with choice E.

---------------------------------

C)

Choice C is false. The max for [tex]h(\theta)[/tex] is at y = 27

---------------------------------

D)

To figure out the amplitude, we can note the vertical distance from the midline y = 14 to the max is 27-14 = 13 units.

Or we can note the distance from the midline to the min is 14-1 = 13 units.

Or we subtract the min and max, and divide by 2:  (yMax-yMin)/2 = (27-1)/2 = 26/2 = 13

Either way, the amplitude is 13 units

Choice D is false.

---------------------------------

E)

Choice E is true. Refer to choice B to see why this is the case.

---------------------------------

Side note: one possible function is [tex]h(\theta) = 13\cos(4\theta+\pi) + 14[/tex]. The various pieces of that function (in)directly lead to the amplitude, period, and vertical shift.

The correct two options are h(Ф) has vertical shift of 14 and Choice E

The period of a fuction is x/2

A function is defined as the expression that set up the relationship between the dependent variable and independent variable.

Let's go through the answer choices

A)

The lowest point is y = 1, and the highest point is y = 27. Computing the midpoint gets us (yMin+yMax)/2 = (1+27)/2 = 28/2 = 14.

The midline of  is y = 14.

Consider that the parent function  has a midline of y = 0. The jump from y = 0 to y = 14 must mean  has a vertical shift of 14.

Choice A is true

B)

The period is the length of the cycle. For the function , the function starts at  and repeats when it gets to . This is a difference of pi/2 units which is the period.

Choice B is false. It contradicts with choice E.

C)

Choice C is false. The max for  is at y = 27

D)

To figure out the amplitude, we can note the vertical distance from the midline y = 14 to the max is 27-14 = 13 units.

Or we can note the distance from the midline to the min is 14-1 = 13 units.

Or we subtract the min and max, and divide by 2:  (yMax-yMin)/2 = (27-1)/2 = 26/2 = 13

Either way, the amplitude is 13 units

Choice D is false.

E)

Choice E is true. Refer to choice B to see why this is the case.

Side note: one possible function is h(Ф) = 13 Cos(4Ф + π) + 14 . The various pieces of that function (in)directly lead to the amplitude, period, and vertical shift.

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[tex]\longrightarrow{\green{ (a + b)(a - b) }}[/tex] 

[tex]\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:EXPLANATION:}}}[/tex]

[tex] ={a}^{2} - {b}^{2} [/tex]

[tex] = (a + b)(a - b)[/tex]

[tex]\bold{ \green{ \star{ \orange{Mystique35}}}}⋆[/tex]

Answer:

[tex]cot\theta=-\frac{5}{12}[/tex]

Step-by-step explanation:

We are given that the point (5,-12) lies on the terminal side of an angle theta in standard position.

We have to find the exact value of [tex]cot\theta[/tex].

Let

Point (x,y)=(5,-12)

[tex]r=\sqrt{x^2+y^2}[/tex]

Substitute the values

[tex]r=\sqrt{5^2+(-12)^2}[/tex]

[tex]r=\sqrt{169}=13[/tex]

We know that

[tex]cot\theta=\frac{x}{y}[/tex]

Using values

[tex]cot\theta=-\frac{5}{12}[/tex]

1 hour is 60 minutes

6 hours x 60 = 360 total minutes

Divide total time by number of lessons:

360/8 = 45

Each lesson was 45 minutes

Answer:

B

Step-by-step explanation:

we rule out all other 3 by the simple fact that the side is 9

Answer:

yes

Step-by-step explanation:

51 divided by 25

Add all the coins to get total coins:

6 + 8 + 12 + 3 = 29 total coins

Subtract dimes to find total of the coins that are not dimes:

29 -6 = 23

Probability of not picking a dime is the number of coins that aren’t dimes over total coins:

23/29

Answer:

2 and -1

Step-by-step explanation:

sum = 2 + (-1)

= 1

product = 2 × ( -1)

= -2

Answer:

y = 360 (1.03)^x

Step-by-step explanation:

Amount in account = principle * (1+ rate) ^ time

When the interest is annual

rate is in decimal form and time is in years

y = 360 ( 1+.03) ^x

y = 360 (1.03)^x

Answer:

The answer is "35.22".

Step-by-step explanation:

Let the given equation:

[tex]\to 130(x-38) +950 \% \times 38=0\\\\\to 130x- 4940+\frac{950}{100} \times 38=0\\\\\to 130x- 4940+9.5\times 38=0\\\\\to 130x- 4940+361=0\\\\\to 130x- 4579=0\\\\\to 130x= 4579\\\\\to x= \frac{4579}{130}\\\\\to x= 35.22[/tex]

Answer: $25

Step-by-step explanation: took the quiz

Answer:

[tex]a = 18x[/tex]

Step-by-step explanation:

Given

[tex]\frac{1}{2}(a + d) = 10x[/tex]

[tex]\frac{1}{2}(b + c) = 8x[/tex]

[tex]\frac{1}{3}(b + c+d) = 6x[/tex]

Required

The value of (a)

We have:

[tex]\frac{1}{2}(a + d) = 10x[/tex] --- multiply by 2

[tex]\frac{1}{2}(b + c) = 8x[/tex] --- multiply by 2

[tex]\frac{1}{3}(b + c+d) = 6x[/tex] --- multiply by 3

So, we have:

[tex]\frac{1}{2}(a + d) = 10x[/tex]

[tex]a + d = 20x[/tex]

[tex]\frac{1}{2}(b + c) = 8x[/tex]

[tex]b + c = 16x[/tex]

[tex]\frac{1}{3}(b + c+d) = 6x[/tex]

[tex]b + c + d= 18x[/tex]

Substitute [tex]b + c = 16x[/tex] in [tex]b + c + d= 18x[/tex]

[tex]16x + d = 18x[/tex]

Solve for d

[tex]d = 18x - 16x[/tex]

[tex]d = 2x[/tex]

Substitute [tex]d = 2x[/tex] in [tex]a + d = 20x[/tex]

[tex]a + 2x = 20x[/tex]

Solve for (a)

[tex]a = 20x - 2x[/tex]

[tex]a = 18x[/tex]

Part one:

[tex]x^2-8x=-26[/tex]

Rewrite in the form [tex](x+a)^{2} =b[/tex]

[tex]\left(x-4\right)^2=-10[/tex]

[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]

Solve [tex]x-4=\sqrt{-10} : x=\sqrt{10} i+4[/tex]

Solve [tex]x-4=\sqrt{-10} : x=-\sqrt{10} i+4[/tex]

[tex]x=\sqrt{10}i+4,\:x=-\sqrt{10}i+4[/tex]

Part two:

[tex]x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:26}}{2\cdot \:1}[/tex]

Simplify [tex]\sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:26}}: 2\sqrt{10} i[/tex]

[tex]=\frac{-\left(-8\right)\pm \:2\sqrt{10}i}{2\cdot \:1}[/tex]

Separate solutions

[tex]x_1=\frac{-\left(-8\right)+2\sqrt{10}i}{2\cdot \:1},\:x_2=\frac{-\left(-8\right)-2\sqrt{10}i}{2\cdot \:1}[/tex]

[tex]\frac{-(-8)+2\sqrt{10}i }{2*1} :4+\sqrt{10}i[/tex]

[tex]\frac{-(-8)+2\sqrt{10}i }{2*1} :4-\sqrt{10}i[/tex]

[tex]x=4+\sqrt{10}i,\:x=4-\sqrt{10}i[/tex]

The solutions are:-

[tex]x=4+\sqrt{10i}\\\\x=4-\sqrt{10i}[/tex]

What is the equation?

The definition of an equation is a mathematical statement that shows that two mathematical expressions are equal.

Here given equation is

[tex]x^2- 8x + 26 = 0\\\\x^2-8x=-26\\\\(x-4)^2=-10\\\\[/tex]

[tex](x-4)=[/tex]±[tex]\sqrt{-10}[/tex]

[tex]x=\sqrt{10}i+4\\\\x=-\sqrt{10}i+4[/tex]

So,

[tex]x=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\\\x=\frac{8+-\sqrt{(-8)^2-4(1)(26)}}{2(1)}\\\\x=\frac{8+-\sqrt{(-8)^2-4(1)(26)}}{2(1)}\\\\x=\frac{8+-\sqrt{64-104}}{2}\\\\x=\frac{8+-2\sqrt{10i}}{2}\\\\x=\frac{2(4+-\sqrt{10i})}{2}\\\\x=4+\sqrt{10i}\\\\x=4-\sqrt{10i}[/tex]

Hence, the solutions are:-

[tex]x=4+\sqrt{10i}\\\\x=4-\sqrt{10i}[/tex]

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