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3 years ago

Solve the equation for x.

1 answer:

3 0

A.

Use PEMDAS

to get rid of the 3/4, divide both sides by 3/4 (.75) and 30 becomes 40

12x + 4 = 40

Subtract 4 from each side to cancel the - 4

12x = 36

Divide both sides by 12 to get rid of the 12

X = 3

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The sum of three consecutive even numbers is 144. What is the smallest of the three numbers ?

ikadub [295]

Answer:

Step-by-step explanation:

132+10+2=144

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4 years ago

WILL GIVE BRAINLIEST& 15 PTS.

kicyunya [14]

Answer:

The given fraction $\frac{x^3-x^2}{x^3}$ reduces to $\frac{x-1}{x}$

Step-by-step explanation:

Consider the given fraction $\frac{x^3-x^2}{x^3}$

We have to reduce the fraction to the lowest terms.

Consider numerator $x^3-x^2$

We can take x² common from both the term,

Thus, numerator can be written as $x^2(x-1)$

Given expression can be rewritten as ,

$\frac{x^3-x^2}{x^3}=\frac{x^2(x-1)}{x^3}$

We can now cancel $x^2$ from both numerator and denominator,

$\Rightarrow \frac{x^2(x-1)}{x^3}=\frac{x^2(x-1)}{x^2 \cdot x}$

$\Rightarrow \frac{(x-1)}{x^2 \cdot x}=\frac{x-1}{x}$

Thus, the given fraction $\frac{x^3-x^2}{x^3}$ reduces to $\frac{x-1}{x}$

6 0

3 years ago

Read 2 more answers

A dog chases a squirrel. The dog is originally 200 feet away from the squirrel. The dog's speed is 150 feet per minute. The squi

Dafna11 [192]

Answer:

15 seconds

Step-by-step explanation:

If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:

d = rt

d+200 = 150t

And the squirrel's:

d = 100t

If we solve both for t and set them equal to each other we have:

$\frac{150}{d+200} =\frac{100}{d}$

Now we can cross multiply to solve for d:

150d = 100d + 20,000 and

50d = 20,000

d = 400

But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:

$t=\frac{100}{d}=\frac{100}{400} =\frac{1}{4}$