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3 years ago

What is the volume of a rectangular prism with a 4 inch base and a height of 4?

Mathematics

1 answer:

timofeeve [1]3 years ago

8 0

V=lwh
V= (4)(4)(4)
V= 64 in cubed
$v = lwh$

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8 × ( 10 - 7 ) distribute property<br><br>

icang [17]

Answer:

8 x (10-7)

80-56

Let me know if this helps!

6 0

2 years ago

5×5×5×5 as an exponent

UNO [17]

5^4 power. We can identify this because 5 is multiplying itself 4 times.

Hope this helps!

5 0

3 years ago

Read 2 more answers

Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question.

weqwewe [10]

Answer:

8(15-m)

Step-by-step explanation:

By looking at the options, we can understand that the right one is 8(15-m), since we need to multiply the time by the number of question.

Also, we can check that result by arranging an equations usint the total time.

55 = 3m - 8(15-m)

55 = 3m - 120 + 8m

55 = 11m - 120

55 + 120 = 11m

175 = 11m

m = 175/11 = 15.91

55 = 3*15.91 - 8(15-15.91) = 47.73 - 120 + 127.28 = 55.01 ≈ 55

6 0

2 years ago

1. The sum of the angles is 90° The difference between twice the larger angle and the smaller

Ostrovityanka [42]

Answer:

$\displaystyle 25°\:and\:°65$

Step-by-step explanation:

{90° = m∠1 + m∠2

{105° = 2m∠1 - m∠2

_______________

$\displaystyle \frac{195°}{3} = \frac{3m∠1}{3} \\ \\ 65° = m∠1$

All you have left to do is this:

$\displaystyle 25° = -65° + 90°$

So, your two angle measures are $\displaystyle 25°\:and\:°65.$

I am joyous to assist you anytime.

5 0

3 years ago

An equation parallel and perpendicular to 4x+5y=19

UNO [17]

Answer:

Parallel line:

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

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

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

$4x+5y=19$

we can change it into y=mx+b form

$5y=-4x+19$

$y=-\frac{4}{5}x+\frac{19}{5}$

so,

$m=-\frac{4}{5}$

Parallel line:

we know that slope of two parallel lines are always same

so,

$m'=-\frac{4}{5}$

Let's assume parallel line passes through (1,1)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-1=-\frac{4}{5}(x-1)$

now, we can solve for y

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

we know that slope of perpendicular line is -1/m

so, we get slope as

$m'=\frac{5}{4}$

Let's assume perpendicular line passes through (2,2)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-2=\frac{5}{4}(x-2)$

now, we can solve for y

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

4 0

3 years ago