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kifflom [539]
3 years ago
8

What is the justification for each step in solving the inequality? −7(3+8x)+5x≤1−62x

Mathematics
2 answers:
lyudmila [28]3 years ago
8 0

Answer:

Step-by-step explanation:

dangina [55]3 years ago
5 0
Distribute -7 because of math
-21-56x+5x≤1-62x
add like terms because you aint chaingin nothing
-21-51x≤1-62x
add 62x to both sides (addition property of equality, where a=a and b=b, a+b=a+b) and also (addative inverse, a+(-a)=0) that turns the -62x to 0
-21+11x≤1
add 21 to both sides (additiona proerty of equality and addativeinverse)
11x≤22
divide both sides by 11 (division property of equality, if a=a and b=b, then a/b=a/b)
x≤2
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Simplify X 1/2 times X 2/3
vovikov84 [41]

Answer:

3/6x • 4/6x

12/6x

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Step-by-step explanation:

you would change both fractions to have the same denominator then simplify if needed

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At summer camp there are 52 boys, 47 girls,and 18 adults.How many people are at summer camp?
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Consider the function f(x) = x2 + 3x − 7. Find the value of f(x) when x = 3 and select the correct answer below.
victus00 [196]

<em>Answer</em>

= 11

<em>Explanation</em>

f(x) = x² + 3x- 7

To get the value of f(x) when x = 3, we substitute x with 3 in the function f(x).

f(x) = x² + 3x- 7

= 3² + (3×3) -7

= 9 + 9 - 7

= 11

6 0
3 years ago
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The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
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