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2 years ago

HERE ARE 2 QUESTIONS PLS HELP ASAP ITS MATH

Mathematics

1 answer:

leva [86]2 years ago

7 0

<h2><u>Answer:</u></h2><h3>6.) $\boxed{A}$ </h3><h3>7.) $\boxed{D}$ </h3><h2 /><h2><u>Solution Steps:</u></h2>

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<h3>6.) Solve each problem:</h3>

$3$ ÷ $8=0.375$

$4$ ÷ $12=0.333333333$

$2$ ÷ $6=0.333333333$

$10$ ÷ $30=0.333333333$

- So looking at this most of them equal 0.333333333, which in fraction form is $\frac{1}{3}$ , so 3 ÷ 8 must be the answer.

<h3 /><h3>7.) Solve each problem: </h3>

$25$ ÷ $8=3.125$

$\frac{25}{8}=3.125$

$3\frac{1}{8}=\frac{25}{8}=3.125$

$8$ ÷ $25=0.32$

<em> - So looking at each problem answer, most of them are 3.125, except for 8 ÷ 25.</em>

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Find the area of the regular polygon. Round your answer to the nearest tenth. 344.3 in.2 172.2 in.2 84.4 in.2 50.0 in.2

My name is Ann [436]

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2 years ago

HELP PLEASE !!!!!!!!!!

Arada [10]

Answer:

Answer B $(-\infty,1)U(1,2]$

Step-by-step explanation:

Notice that the quotient of f(x)/g(x) is:

$h(x)=\frac{\sqrt{2-x} }{x-1}$

therefore, this new function imposes conditions due to the fact that it has a square root in the numerator and a binomial in the denominator both with the variable x. Then, in order for the root in the numerator to be defined, the argument inside the root must be larger than or equal to zero. That is:

$2-x\geq 0\\2\geq x\\x\leq 2$

So, this condition must be satisfied by the x-values of the domain.

Then we have the binomial in the denominator, which in order to be defined needs to be different from zero. Notice that the only x-value that could cause problems (render zero) is:

$x-1=0\\x=1$