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3 years ago

50 POINTS + BRAINLIEST PLEASE HELP

Mathematics

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

The slope is 4/5

Step-by-step explanation:

The line goes over the point (5,6) so we can use that to help us find the slope. Since slope is rise/run, the rise would be 4 and the run would be 5. Thus the slope is 4/5

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A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is

Illusion [34]

Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume $V$ of a solid pyramid can be given as:

$\displaystyle V = \frac{1}{3} \cdot b \cdot h$ ,

where

$b$ is the area of the base of the pyramid, and
$h$ is the height of the pyramid.

Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be $x$ cm.

At the vertice of this pyramid, $x = 0$ and the length of a side of the square is also $0$ .
At the base of this pyramid, $x = 3$ and the length of a side of the square is $4$ cm.

As a result, the length of a side of the square will be

$\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x$ .

The area of the square will be

$\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}$ .

Integrate the area of the horizontal cross-section with respect to $x$

from the top of the pyramid, where $x = 0$ ,
to the base, where $x = 3$ .

$\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}$ .

In other words, the volume of this pyramid is 16 cubic centimeters.

5 0

3 years ago

2. How far is the bird from the worm? Round to the nearest whole number if necessary. (TEKS

Vikentia [17]

A)60 ft
I pretty sure

4 0

3 years ago

Read 2 more answers

What is the solution to this system of equations?

antoniya [11.8K]

Answer:

Step-by-step explanation:

Since both equations express y in terms of x we can equate the right sides

3x + 5 = - x + 3 ( add x to both sides )

4x + 5 = 3 ( subtract 5 from both sides )

4x = - 2 ( divide both sides by 4 )

x = - 0.5

substitute x = - 0.5 in either of the 2 equations

using y = - x + 3 then y = 0.5 + 3 = 3.5

solution is (- 0.5, 3.5 ) → c

6 0

3 years ago

D 2.<br> The sum of 3x2 – 8x – 2 and 4x – 2 is

maw [93]

Answer:

3x^2 -4x -4

Step-by-step explanation:

3x^2 – 8x – 2 + 4x – 2

Combine like terms

3x^2 -8x+4x -2 -2

3x^2 -4x -4

8 0

4 years ago

Read 2 more answers

Air pressure may be represented as a function of height (in meters) above the

denis-greek [22]

Answer: OPTION C.

Step-by-step explanation:

<h3> The complete exercise is: "Air pressure may be represented as a function of height (in meters) above the surface of the Earth, as shown below:</h3><h3> $P(h) =P_0e^{-0.00012h}$ </h3><h3> In this function $P_o$ is the air pressure at the surface of the earth, and $h$ is the height above the surface of the Earth, measured in meters. At what height will the air pressure equal 50% of the air pressure at the surface of the Earth"</h3><h3></h3>

Given the following function:

$P(h) =P_0e^{-0.00012h}$

In order to calculate at what height the air pressure will be equal 50% of the air pressure at the surface of the Earth, you can follow these steps:

1. You need to substitute $P(h)=0.5P_o$ into the function:

$0.5P_o=P_0e^{-0.00012h}$

2. Finally, you must solve for $h$ .

Remember the following property of logarithms:

$ln(b)^a=a*ln(b)\\\\ln(e)=1$

Then, you get this result:

$0.5P_o=\frac{P_o}{e^{0.00012h}}\\\\(0.5P_o)(e^{0.00012h})=P_o\\\\e^{0.00012h}=\frac{P_o}{0.5P_o}\\\\ln(e)^{0.00012h}=ln(2)\\\\0.00012h*1=ln(2)\\\\h=\frac{ln(2)}{0.00012}\\\\h=5576.2$

7 0

4 years ago