Can someone help me please tho, nobody answered this and I don’t see it on brainly neither..

Use the percent bar model to find the missing percent

Ne4ueva [31]

A percentage is a way to describe a part of a whole. The missing percent on the bar model is equal to 12.5%.

<h3>What are Percentages?</h3>

A percentage is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25 which is equal to 25%.

To convert a fraction to a percentage, convert the fraction to decimal form and then multiply by 100 with the '%' symbol.

On the percent bar model, above the missing percentage, the fraction is given as 1/8.

Now, the given fraction 1/8 in the form of percentage can be written as,

1/8

= 1/8 × 100%

= 0.125 × 100%

= 12.5%

Hence, the missing percent on the bar model is equal to 12.5%.

Learn more about Percentages here:

brainly.com/question/6972121

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3 0

1 year ago

Simplify. (3m^2n4)^3 <br>A)27m^5n^7 <br>B)9m^6n12 <br>C)27m^6n^12<br>D)9m^5n^7

butalik [34]

(3m²n⁴)³ = 27m^6n^12. choice C

6 0

3 years ago

What is three more than the product of a number and 7 equals 8 . use the variable x for the unknown number?

Anastaziya [24]

(7x) + 3 = 8
subtract both sides by 3.

7x = 5
divide both sides by 7 to isolate the variable, x.

x = 5/7

7 0

2 years ago

Plsssss helpp in the next 2 min gives 10 points

saul85 [17]

Answer:

The answer is 144,5

Step-by-step explanation:

144,5/170×100 =85%

4 0

2 years ago

Read 2 more answers

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

2 years ago