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2 years ago

9

Solve the following inequality algebraically.

xFormula1" title=" |x - 1| \leqslant 13" alt=" |x - 1| \leqslant 13" align="absmiddle" class="latex-formula">

1 answer:

olya-2409 [2.1K]2 years ago

8 0

lx-1 l ≤ 13

There are 2 solutions:

x-1 ≤ 13

and

x-1 ≥ -13

Solve each one

x-1 ≤ 13

add 1 to both sides of the equation

x-1+1≤ 13+1

x ≤ 14

x-1 ≥ -13

Add 1 to both sides:

x-1+1 ≥-13+1

x ≥ -12

-12 ≤ x ≤ 14

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Answer:

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Step-by-step explanation:

Given:

Output 2009 = 1.5 million dollar

Projected decrease output = 5%

Find:

Output 2019

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3 years ago

Ginny's piggybank contained $8.80 in quarters, dimes, and nickels. there were two more than five times as many nickels as quarte

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Answer:

There are 13 quarters, 67 nickels, and 22 dimes.

Step-by-step explanation:

The value of a quarter is $0.25, the value of dime is $0.10, and the value of a nickel is $0.05. Let there be q quarters, n nickels, and d dimes.

Let's write equations from the statements given:

<u>"Ginny's piggybank contained $8.80 in quarters, dimes, and nickels.":</u>

$0.25q+0.1d+0.05n=8.80$

<u>"there were two more than five times as many nickels as quarters":</u>

$5q+2=n$

<u>"4 less than twice as many dimes as quarters":</u>

<u> $2q-4=d$ </u>

Plugging in equation 2 and 3 into equation 1, we can solve for q:

$0.25q+0.1(2q-4)+0.05(5q+2)=8.80\\0.25q+0.2q-0.4+0.25q+0.1=8.80\\0.7q=8.80+0.4-0.1\\0.7q=9.1\\q=\frac{9.1}{0.7}=13$

<em>Substituting q = 13 into equation 2 will give us nickels:</em>

$n=5q+2\\n=5(13)+2\\n=67$

<em>Substituting q = 13 into equation 3 will give us dimes:</em>

$d=2q-4\\d=2(13)-4\\d=22$

There are 13 quarters, 67 nickels, and 22 dimes.

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