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3 years ago

Can somebody help me with this?

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

6 0

Answer:

E) 1/2

Step-by-step explanation:

This is no ordinary problem. It actually requires you to solve two different problems in order to figure out the correct answer.

First, we must figure out what the hypotenuse of the triangle is. We have the two small sides (5 and 12), but don't have the larger side. To find this, we must use the Pythagorean theorem: $a^{2} + b^2 =c^2$

Let's plug in our a and b values

$5^2 + 12^2 = C^2$

Simplify

$144 + 25 = c^2$

$169 = c^2$

Square both sides

$c = 13$

But sadly, we aren't done yet. Now we know the length of the whole hypotenuse side. Time to figure out what X is by a simple equation.

$12x + 14x = 13$

Simplify

$26x = 13$

Divide by 26

Answer is 1/2

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A multiple regression model has ___

Salsk061 [2.6K]

Answer:

C. More than one independent variable

Step-by-step explanation:

The difference between linear regression model and multiple regression model is that the linear regression has just one independent variable which is use to determine the dependent variable. While the multiple regression model has at least two or more independent variable which is use to determine the dependent variable.

5 0

3 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

A student used a 30% off coupon to buy a jacket. There was no tax. The student paid $67.34. What was the original price of the j

PSYCHO15rus [73]

Step-by-step explanation:

the solution is in the picture

6 0

3 years ago

Read 2 more answers

Rectangle A has a length of 2x + 6 and a width of 3x. Rectangle B has a length of x + 2 and an area of 12 square units greater t

Maurinko [17]

So here is how you solve for the answer.
Firstly, you solve for the Area of Rectangle A.
The formula for Area is Length x width.
So A = (2x + 6)(3x) and the result is: 6x^2 + 18x
Now, let y be the width of rectangle B.
<span>(x+2) (y) = 6x^2 + 18x + 12
(x+2) y = 6(x+1)(x+2)
y = 6(x+1)
</span>So the final answer would be width is 6x + 6. The answer is the third option. Hope this answer helps.

7 0

4 years ago

Eliminate the parameter. x = t2 + 2, y = t2 - 4

stepan [7]

$\bf \begin{cases}
x=t^2+2\implies x-2=t^2\implies \sqrt{x-2}=\boxed{t}\\\\
y=t^2-4\\
----------\\
y=\left( \boxed{\sqrt{x-2}} \right)^2-4\implies y=x-2-4\\\\
y=x-6
\end{cases}$

just a plain vanilla substitution

8 0

3 years ago

Read 2 more answers