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3 years ago

What is the slope for 10,15 and 15,20

Mathematics

1 answer:

makvit [3.9K]3 years ago

6 0

So it will be y1-y2, so 15-20 is -5

Then x1-x2, so 10-15 is -5.

-5/-5 = 1

There’s ya slope : 1

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Please help me :)) algebra 2! what are the real roots?

irina1246 [14]

Answer:

Step-by-step explanation:

So we have the equation:

$3(x-4)^\frac{4}{3}+16=64$

First, let's subtract 16 from both sides:

$3(x-4)^\frac{4}{3}=48$

Now, let's divide both sides by 3:

$(x-4)^\frac{4}{3}=16$

Remember that with fractional exponents, we can move the denominator into the root position. Therefore:

$(\sqrt[3]{x-4})^4=16$

Let's take the fourth root of both sides. Since we're taking an even root, make sure to have the plus-minus symbol!

$\sqrt[3]{x-4} =\pm 2$

Cube both sides. Since we're cubing, the plus-minus stays.

$x-4=\pm 8$

Add 4 to both sides.

$x=\pm 8+4$

Calculator:

$x=8+4\text{ or } x=-8+4\\x=12\text{ or } x=-4$

So, our answer is A.

And we're done!

7 0

3 years ago

The school gym can hold 650 people. A permanent bleacher in the gymn holds 136 people. The event organizers are setting up 25 ro

igomit [66]

650 -136=514
514 divided by 25 = 20.56
So it would be 21
This will give you a little extra chairs but they have to be equal in each row

3 0

3 years ago

WILL MARK YOU BRAINLIEST

anzhelika [568]

Answer:The area is 14306. 625 cm.

Step-by-step explanation:

To find the area of the circle, we will need to find the diameter and find the radius.

So we know the circumference is 423.9 m and we know to get the circumference, you will need to multiply pie by the diameter .

so now we need divide the circumference by pie to find the diameter.

423.9/ 3.14 = 135 so 135 is the diameter.

To find the radius we need to divide the diameter by 2.

135/2 = 67.5

Now is time to find the area which uses the formula A= nr^2

A = 3.14 * 67.5^2

A = 14306. 625

5 0

3 years ago

Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

2 years ago

What is the absolute value of the square root of 5 minus 5?

OLEGan [10]

-2.7639320225 is the exact when estimated -3 hope this helped if can pls mark as brainliest have a great day!

6 0

3 years ago