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Misha Larkins [42]

3 years ago

12

15+ Points for BEST Answer!

1 answer:

Stells [14]3 years ago

6 0

Answer: 306in^3

method: I would split the cross-section (the trapezium at the front) into two shapes. You first have your rectangle with dimensions of 4 and 7, and your triangle with dimensions of height 4 and base 3 (10 - 7 = 3). 4 × 7 = 28in^2, 4 × 3 = 12, 12 ÷ 2 = 6in^2. The total surface area of the cross-section is therefore 6 + 28 = 34in^2. Now, to find the volume, you just multiply that by 9 which is 306in^3, which is your answer. I hope this helps!

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WITCHER [35]

$72.31 is the total amount AFTER ADDING TWO DEPOSITS

So, if we want to know the amount before deposits, we just need to take the total of previous two deposits

The expressions that show your balance would be

$72.31 – $17.92 – $15.33; $39.06

6 0

3 years ago

Select the two values of x that are roots of this equation.

RSB [31]

The answers are A and B

7 0

2 years ago

What is the x value of the triangle 3x-5, 2x+20, 4x-30

docker41 [41]

<u>Answer:</u>

x = 21.7°

<u>Step-by-step explanation:</u>

We know that the sum of the three angles in triangle add up to a total of 180°.

So adding the given expressions of angles and putting them equal to 180 to find the value of x:

$(3x-5) + (2x+20) +(4x-30)= 180$

$4x+3x+2x+20-30-5=180$

$9x=180+30-20+5$

$9x=195$

$x=\frac{195}{9}$

$x=21.7$

Therefore, the value of x in a triangle with angles 3x-5, 2x+20, 4x-30 is equal to 21.7°.

4 0

3 years ago

Read 2 more answers

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atroni [7]

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Step-by-step explanation:

3 0

3 years ago

Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear

AlladinOne [14]

Answer:

$x=\frac{83}{50}$

Step-by-step explanation:

we know that

If the three points are collinear

then

$m_A_B=m_A_C$

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope AB

we have

$A(1,\frac{2}{3}),B(x,-\frac{4}{5})$

substitute in the formula

$m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}$

$m_A_B=\frac{\frac{-12-10}{15}}{x-1}$

$m_A_B=-\frac{22}{15(x-1)}$

step 2

Find the slope AC

we have

$A(1,\frac{2}{3}),C(-\frac{1}{2},4)$

substitute in the formula

$m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}$

$m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}$

$m_A_C=-\frac{20}{9}$

step 3

Equate the slopes

$m_A_B=m_A_C$

$-\frac{22}{15(x-1)}=-\frac{20}{9}$

solve for x

$15(x-1)20=22(9)$

$300x-300=198$

$300x=198+300$

$300x=498$

$x=\frac{498}{300}$

simplify

$x=\frac{83}{50}$

8 0

4 years ago