All categories

3 years ago

Please answer this question!! 20 points and brainliest!

Mathematics

1 answer:

Charra [1.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

We can find the surface area of the object by adding the surface areas of each part. We have many rectangle faces to count and two triangular faces. Each has a formula for the area. We will find the area of each and then add them all together.

Triangle - 0.5 *b*h

Rectangle - b*h

Triangles

There are two triangles on either side. The height is 1.5. The base is 1.8.

0.5(1.5)(1.8)=1.35 meters squared

Since there are two, we will add 1.35+1.35 in our final calculation.

Rectangles

We will start by calculating the largest rectangle on the side. It has height of 4 and a base of 2.5 (shown above left).

4(2.5)=10

Since there are two (one we can see and one we can't), we will add 10+10 in our final calculation.

Next we calculate the top and bottom. The height is 3 and the base is 2.5 on top. But the bottom sticks out more and adds 1.8 to its base.

Top - 3(2.5)=7.5

Bottom-3(2.5+1.8)=12.9

Finally, we will calculate the front side and back(not visible) as well as the slant up front. The back side has height 4 and base 3. The front side has base 3 and height 4-1.5=2.5. The slant has base 2.3 and height 3.

Back - 4(3)=12

Front- 3(2.5)=7.5

Slant - 3(2.3)=6.9

We add all together for the total surface area: 1.35+1.35+10+10+7.5+12.9+12+7.5+6.9=69.5 meters squared.

You might be interested in

If you awnser this correctly I will give you more points

Scorpion4ik [409]

Answer:

1) 5m-10

Parts, 5m, -10

Operation, Subtraction

Word Expression, 10 less than the product of 5 and m.

Step-by-step explanation:

1) 5m-10

Parts, 5m, -10

Operation, Subtraction

Word Expression, 10 less than the product of 5 and m.

2) 12/2+7

6+7

3) 8y-(22)

Terms, 8y, 22

Operation, Subtraction

Word Expression, 22 less than the product of 8 and y.

8 0

3 years ago

Read 2 more answers

If rhe discriminant of a quadratic equation is equal to -8 which statement describes the roots

Nastasia [14]

Answer:

Step-by-step explanation:

Given

$Discriminant = -8$

Required

Determine the type of roots

Represent Discriminant with D; such that

$D = -8$

D is calculated as thus

$D = b^2 - 4ac$

And it has the following sequence of results

When $D > 0;$ then the roots of the quadratic equation are real but not equal

When $D = 0;$ then the roots of the quadratic equation are real and equal

When $D < 0;$ then the roots of the quadratic equation are complex or imaginary

Given that $D = -8$ ; This means that $D < 0;$ and base on the above analysis, we can conclude that the roots of the quadratic equation are complex or imaginary

6 0

3 years ago

Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A What is the value of the A^2?

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$