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makkiz [27]
3 years ago
11

Solve the following inequality algebraically. ∣x−5∣≤11

Mathematics
1 answer:
Serga [27]3 years ago
7 0

Answer:

-6 ≤ x ≤ 16

Step-by-step explanation:

First, apply absolute value rule:

x - 5 ≤ 11 and x - 5 ≥ -11

Then, we add 5 on both sides with both inequalities:

x ≤ 16 and x ≥ -6

Finally, combine the inequalities:

-6 ≤ x ≤ 16

And we're done ^^

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Dee Dee bought an Apple for $0. 85, a sandwich for $4. 50, and a bottle of water for $1. 50. How much did Dee Dee spend?
OLga [1]

To figure out how much she spent all we have to do is add up $0.85, $4.50 and $1.50

$0.85 + $4.50 + $1.50 = $6.85

Therefore she spent $6.85

Hope this helps

8 0
3 years ago
Read 2 more answers
(x+40)(x+40) show steps
pishuonlain [190]

(x + 40) * (x + 40) =

(x + 40)² =

x² + 80x + 1600

6 0
3 years ago
What is the exact solution to the equation 4^5x=3^(x-2)<br> list the steps to get the answer
scoray [572]

Answer: x = -0.377

Step-by-step explanation:

We have the equation:

4^(5*x) = 3^(x - 2)

Now we can use the fact that:

Ln(A^x) = x*Ln(A)

Then we can apply Ln(.) to both sides of the equation to get:

Ln(4^(5*x)) = Ln(3^(x - 2))

(5*x)*Ln(4) = (x - 2)*Ln(3)

(5*x)*Ln(4) - x*Ln(3) = -2*Ln(3)

x*(5*Ln(4) - Ln(3)) = -2*Ln(3)

x = -2*Ln(3)/(5*Ln(4) - Ln(3)) = -0.377

6 0
3 years ago
Jim ran 15 miles in 5 days. At this rate, how many days will it take Jim to run 165 miles?
AleksAgata [21]

Answer:

55 days

Step-by-step explanation:

Given

Jim ran 15 miles in 5 days

no. of miles ran in 5 days = 15 miles

dividing LHS and RHS by 5

no. of miles ran in 5/5(=1) days = 15/5 miles = 3 miles

no. of miles ran in 1 day =  3 miles

let the no. of days taken to run 165 miles be x   ----A

No of miles ran in x days = x*no. of miles ran in 1 day = 3x miles

thus,  From A

3x  = 165

x = 165/3 = 55

Thus, it took 55 days for JIM to run 165 miles

3 0
4 years ago
What is the difference? startfraction 2 x 5 over x squared minus 3 x endfraction minus startfraction 3 x 5 over x cubed minus 9
weeeeeb [17]

The difference of the expression \frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x} is \frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}

<h3>How to determine the difference?</h3>

The expression is given as:

\frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x}

Factor the denominators of the expressions:

\frac{2x^5}{x(x - 3)} - \frac{3x^5}{x(x^2 - 9)}

Apply the difference of two squares to x² - 9

\frac{2x^5}{x(x - 3)} - \frac{3x^5}{x(x - 3)(x + 3)}

Take LCM

\frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}

Hence, the difference of the expression \frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x} is \frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}

Read more about expressions at:

brainly.com/question/723406

#SPJ4

6 0
2 years ago
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