All categories

2 years ago

4c = d for c Solve for indicated variable.

Mathematics

2 answers:

anzhelika [568]2 years ago

7 0

Answer:

c = d/4

Step-by-step explanation:

solve this literal equation using inverse operations:

4c = d

/4 /4

c = d/4

motikmotik2 years ago

4 0

Answer:

c=1/4d

Step-by-step explanation:

You might be interested in

Solve the inequality. Then graph the solutions on a number line. 20 > - 4(2x - 7)

sweet-ann [11.9K]

Answer:

Step-by-step explanation:

Multiply -4 with 2x and -7

Then collect like terms

Divide both sides by the co-efficient of x

8 0

2 years ago

What are the factors of x^2-64

mel-nik [20]

Difference Of Two Squares
<span>a^2 - b^2 = a^2 - b^2
</span><span>so
x^2-64
=x^2 - 8^2
= (x + 8)(x - 8)</span>

8 0

3 years ago

On average, at least 25,000

dolphi86 [110]

98% are located each day for a total of 24,500 pieces of found luggage, but lost each day would be 500 pieces of luggage.

6 0

2 years ago

I don’t know where to start, I need help.

NeX [460]

Answer:

72 cm^3

Step-by-step explanation:

find the volume of the big block and the volumem of the smaller block and subtract it

(8 x 2 x 5) - (2 x 2 x2)

= 80 cm^3 - 8 cm^3

= 72cm^3

5 0

3 years ago

Find the volume of the solid generated when R (shaded region) is revolved about the given line. x=6−3sec y, x=6, y= π 3, and

Dmitrij [34]

Answer:

$V=9\pi\sqrt{3}$

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

$V=\int\limits^b_a {\pi r^{2}} \, dy$

where:

$r=6-(6-3 sec(y))$

$r=3 sec(y)$

a=0

$b=\frac{\pi}{3}$

so the volume becomes:

$V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy$

This can be simplified to:

$V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy$

and the integral can be rewritten like this:

$V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy$

which is a standard integral so we solve it to:

$V=9\pi[tan y]\limits^\frac{\pi}{3}_0$

so we get:

$V=9\pi[tan \frac{\pi}{3} - tan 0]$

which yields:

$V=9\pi\sqrt{3}$ ]

6 0

3 years ago