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2 years ago

What is the area of cross-section ADGF of this right rectangular prism?

Mathematics

1 answer:

dedylja [7]2 years ago

5 0

Answer:

C.52 square units

Step-by-step explanation:

We are given that

Length of cross-section ADGF,l=13units

Length of cross-section ADGF, b=4 units

We have to find the area of cross-section ADGF of this right rectangular prism.

We know that

Area of rectangle= $l\times b$

Using the formula

Area of cross-section ADGF of this right rectangular prism

= $13\times 4$

Area of cross-section ADGF of this right rectangular prism=52square units

Option C is correct.

C.52 square units

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6 - 2x = 6x = 10 + 6

Natalija [7]

Answer:Simplifying

6 + -2x = 6x + -10x + 6

Reorder the terms:

6 + -2x = 6 + 6x + -10x

Combine like terms: 6x + -10x = -4x

6 + -2x = 6 + -4x

Add '-6' to each side of the equation.

6 + -6 + -2x = 6 + -6 + -4x

Combine like terms: 6 + -6 = 0

0 + -2x = 6 + -6 + -4x

-2x = 6 + -6 + -4x

Combine like terms: 6 + -6 = 0

-2x = 0 + -4x

-2x = -4x

Solving

-2x = -4x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '4x' to each side of the equation.

-2x + 4x = -4x + 4x

Combine like terms: -2x + 4x = 2x

2x = -4x + 4x

Combine like terms: -4x + 4x = 0

2x = 0

Divide each side by '2'.

x = 0

Simplifying

x = 0

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The illustration below shows the graph of y as a function of x.

zloy xaker [14]

Answer:

x = 18

Step-by-step explanation:

5 0

3 years ago

Please help me with this question

adoni [48]

Answer:

$y = e^{x^2 - 3x}\\\\\\Let \ u = x^2 - 3x\\\\y = e^u\\\\\frac{dy}{du} = e^u\\\\\frac{du}{dx} = 2x - 3\\\\y' = \frac{dy}{dx} = \frac{dy}{du} *\frac{du}{dx} = e^u * (2x -3)\\\\Substitute \ u = x^2 -3\\\\y' = (2x -3)e^{x^2-3x}\\\\option A$

5 0

3 years ago

If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2. Find

dimaraw [331]

The formula for average velocity between two times t1 and t2 of the position function f(x) is (f(t2)-f(t1)) / (t2-t1)
Plugging the values in for the first time period we get (f(2.5)-f(2)) / (2.5-2)
=> (f(2.5)-f(2)) / 0.5
f(2) will be the same for all 4 time periods and is
48(2)-16(2)^2 = 32
Now we plugin the other values
f(2.5) = 48(2.5)-16(2.5)^2 = 20
f(2.1) = 48(2.1)-16(2.1)^2 = 30.25
etc.
f(2.05) = 31.16
f(2.01) = 31.8384
Now plug these values into the formula
(20-32)/0.5 = -24
(30.25-32)/0.1 = -17.5
etc.
= -16.8
= -16.16
Final answer:
2.5s => -24 ft/s
2.1s => -17.5 ft/s
2.05 => -16.8 ft/s
2.01 => -16.16 ft/s
Hope I helped :)

6 0

3 years ago

I need this problem solved real quick

PIT_PIT [208]

Let's take this problem step-by-step:

The ;unlabeled angle' adjacent to the 'outside angle of measure 78°'

⇒ is on a straight line

⇒ sum of angle measure = 180 degrees

$unlabeled_.angle+78 = 180\\unlabeled_.angle = 102$

Now we know:

⇒ sum of all angles in a triangle ⇒ 180°

Let's put that in equation form and solve:

$x + 102 + 57=180\\x + 159=180\\x = 21$

Answer: 21°

Hope that helps!

#LearnwithBrainly

8 0

2 years ago