All categories

2 years ago

Im trying to get everything on my overdue list done- but most of its math anyone help?

Mathematics

1 answer:

Bad White [126]2 years ago

3 0

The total surface area of the given pyramid is:

A = 39ft²

<h3>How to get the surface area of the given figure?</h3>

The total surface area will be equal to the sum of the areas of the square and the 4 triangles.

Remember that for a square of side length S, the area is:

A = S².

In this case, S = 3ft, then:

A = (3ft)² = 9ft².

Now, the area of a triangle of base B and height H is:

A = B*H/2.

Here we can see that the triangles have a base of 3ft (the sides of the square) and a height of 5ft, then the area of each triangle is:

A = (3ft)*(5ft)/2 = 7.5 ft²

Then the total area of the figure is:

A = 4*(7.5 ft²) + 9ft² = 39ft²

If you want to learn more about pyramids:

brainly.com/question/10042135

#SPJ1

You might be interested in

He figure is rotated counterclockwise. Which rotation could have taken place?

ruslelena [56]

Answer:

a 90° rotation

I'm not sure about answer...

7 0

3 years ago

Determine the constant of variation for the direct variation given.

ohaa [14]

Answer:

B. 5

Step-by-step explanation:

We have been given that R varies directly with S. When S is 16, R is 80. We are asked to find constant of variation.

We know that two directly proportional quantities are in form $y=kx$ , where,

k = Constant of variation.

Upon substituting our given values, we will get:

$R=k*S$

$80=k*16$

$\frac{80}{16}=\frac{k*16}{16}$

$5=k$

Therefore, the constant of variation is 5 and option B is the correct choice.

6 0

3 years ago

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

3 0

3 years ago

Drag each tile to the correct box. Not all tiles will be used

Illusion [34]

Answer:

$\frac{2(vt-d)}{t^2}=a$

Step-by-step explanation:

Expression to calculate the displacement 'd' is,

d = vt - $\frac{1}{2}at^{2}$

By subtracting vt from both the sides of the equation.

d - vt = - $\frac{1}{2}at^2$

vt - d = $\frac{1}{2}at^{2}$ --------(1)

Multiplying with 2 on both the sides of the equation,

2(vt - d) = at² --------(2)

Dividing by 't²' on both the sides of the equation,

$\frac{2(vt-d)}{t^2}=\frac{at^2}{t^2}$

$\frac{2(vt-d)}{t^2}=a$ --------(3)

Therefore, expression to calculate the acceleration 'a' will be

$\frac{2(vt-d)}{t^2}=a$

4 0

3 years ago

Type the equation. that shows the relationship between the variables in this charty=mx b

ehidna [41]

M=(y-y)/(x-x) That is your answer I hope it helps

3 0

3 years ago