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3 years ago

The GCF of 75,100 and 175 is 25

2 answers:

4 0

TRUE. 25 is the greatest common factor (GCF) of 75,100 and 175. There is no factor higher than 25 that both these numbers share.

Phoenix [80]3 years ago

3 0

Yes the gcf of 75100 and 175 is 25

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What is 1 and 7/10 as a decimal? 0.17 1.7 th 0.7 7.1 1.07

earnstyle [38]

1.7 is the answer to the question

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3 years ago

Please answer this question now

shusha [124]

Hi there! :)

Answer:

$\huge\boxed{V = 359.01 mm^{3} }$

Use the formula V = 1/3(bh) to solve for the volume of the cone where b = πr² where π ≈ 3.14:

Find the area of the base:

b = π(7)²

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V = 1/3(153.86 · 7)

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3 years ago

Help pls and thank you :)

saul85 [17]

D (7) should be the correct answer.

6 0

3 years ago

Read 2 more answers

I need help with my algebra 2. I also want to know how to solve this.

Lisa [10]

Add 1 to both sides:

$\sqrt{x+3} = x+1$

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

$x+1 \geq 0 \iff x \geq -1$

Under this assumption, we square both sides and we have

$x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0$

The solutions to this equation are

$x = -2,\ x=1$

But since we can only accept solutions greater than -1, we discard $x=-2$ and accept $x=1$ .

In fact, we have

$x=-2 \implies \sqrt{-2+3}-1=0\neq -2$

and

$x=1 \implies \sqrt{4}-1=1$

which is the only solution.

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3 years ago

0 A maker of a certain brand of low-fat cereal bars claims that the average saturated fat content is 0.5 gram. In a random sampl

balandron [24]

Answer:

The first step is to determine the average

$x= \frac{0,6+0,7+ 0,7+0,3+0,4+ 0,5+ 0,4 +0,2}{8}\\\ x= 0,475$

The exercise says it’s a normal distribution: (n=8)

$s^{2}= \frac{1}{n-1} ((0,6.x)^2+2(0,7-x)^2+(0.3-x)^2+2(0.4-x)^2+(0.5-x)^2+(0.2-x)^2)\\\ \frac{1}{7}*0,235 = 0,0336 \\\\ s= 0183$

According to the exercise, the mean is equal to 0,5 then the value of t of the distribution can be obtained

$t= \frac{x-u}{\frac{s}{\sqrt{n} }}\\\ t= \frac{0,475 - 0,5}{0,183}\\\ t= -0,3860$

The variable t has 7 grade to liberty, we calculate the p-value as:

$2* P(t < - 0,3860)= \\\ 2* 0,3554 = 0,7108$

This value is very high, therefore the hypothesis is not rejected

7 0

4 years ago