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2 years ago

Which equation is in point-slope form and is depicted by the line in this graph?

Mathematics

1 answer:

Gala2k [10]2 years ago

8 0

C is the answer.........................

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What number must you add to complete the square x^-8x=39

notsponge [240]

Answer:

Given equation: $x^2-8x =39$

when we complete the square , we take half of the value of 8 , then square it, and added to the left sides, we get;

$x^2-8x+4^2 = 39 +4^2$

∵8 is the value $(\frac{8}{2})^2$

Notice that, we add this both sides so that it maintains the equality.

then;

$x^2-8x+4^2 = 39 +4^2$

$(x-4)^2 = 39 + 16$ [ $(a-b)^2 = a^2-2ab+b^2$ ]

Simplify:

$(x-4)^2 =55$

The number must be added to complete the square is, $4^2 = 16$

3 0

3 years ago

Read 2 more answers

(-8) - 3y= 25<br><br>if you have time please help, thank you

Neko [114]

(-8) - 3y= 25

-8-3y= 25

move -8 to the other side

sign changes from -8 to +8

-8+8-3y=25+8

-3y= 33

divide both sides by -3 to get y by itself

and to get +y

-3y/-3= 33/-3

Answer:

y= -11

8 0

2 years ago

Read 2 more answers

Which set of numbers could represent the lengths of the sides of a right triangle?

DiKsa [7]

Answer:

The first set: 8, 15, and 17.

Step-by-step explanation:

By the pythagorean theorem, a triangle is a right triangle if and only if

$\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2$ .

In this case,

$\text{longest side}^2 = 17^2 = 289$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}$ .

In other words, indeed $\text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2$ . Hence, 8, 15, 17 does form a right triangle.

Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two

Factor out the common factor $2$ to simplify the calculations.

$\text{longest side}^2 = 20^2 = 400$

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = 11^2 = 121$ .

<h3> $\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}$ .</h3>

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

6 0

3 years ago

Convert 7.23% into a decimal

Sedaia [141]

Answer:

0.0723

Step-by-step explanation:

To convert 7.23% into a decimal, you just need to divide 7.23 by 100, or just move the decimal point 2 times to the left.

8 0

3 years ago

Read 2 more answers

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par

lyudmila [28]

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

$a(t)=\frac{1}{t}$

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$

5 0

3 years ago