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3 years ago

6

How many times should you multiply 8/7 by itself?

1 answer:

Lana71 [14]3 years ago

6 0

Answer:

to gain what result

Step-by-step explanation:

if u can please the full question

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Write a fraction equivalent to 3/5 that has a denominator of 100.

yuradex [85]

Answer:

60/100

Step-by-step explanation:

We know that 20 goes into 100 5 times, so we multiple numerator and denominator by 20 to get an equivalent fraction:

$\frac{3}{5} (\frac{20}{20} )=\frac{3(20)}{5(20)} =\frac{60}{100}$

6 0

3 years ago

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Round 836 to the hundred

Ksivusya [100]

836 to the nearest hundred is 800 because number under 50 of the last two digits we round it below. Same thing as above 50 we round it up.

4 0

4 years ago

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The set of ordered pairs below represents a linear function. Find the rate of change. {(1,4), (2,6), (3,8), (4,10)} The

Finger [1]

Answer:

(1,2)

Step-by-step explanation:

6 0

4 years ago

Three weeks ago John bought stock at 49 1/4; today the stock is valued at 49 7/8. We could say the stock is performing at which

Bad White [126]

Answer:

The correct option is D.

Step-by-step explanation:

John bought stock at $49\frac{1}{4}$ ; today the stock is valued at $49\frac{7}{8}$ .

$49\frac{1}{4}=49+\frac{1}{4}$

$49\frac{1}{4}=49+0.25$

$49\frac{1}{4}=49.25$

The value of stock at the time of purchasing is 49.25.

$49\frac{7}{8}=49+\frac{7}{8}$

$49\frac{7}{8}=49+0.875$

$49\frac{7}{8}=49.875$

The value of stock at current time is 49.875.

$49.875-49.25=0.625$

The value of stock increased by 0.625.

Since the value of stock increases, therefore the correct option is D, i.e., Above par.

7 0

4 years ago

Read 2 more answers

Which value, when placed in the box, would result in a system of equations with infinitly many solutions? Y=2x - 5 2y-4x=

Nezavi [6.7K]

Given:

The system of equations is

$y=2x-5$

$2y-4x=?$

To find:

The missing value for which the given system of equations have infinitely many solutions.

Solution:

Let the missing value be k.

We have,

$y=2x-5$

$2y-4x=k$

Taking all the terms on the left side, the given equations can be rewritten as

$-2x+y+5=0$

$-4x+2y-k=0$

The system of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinitely many solutions if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

We have,

$a_1=-2,b_1=1,c_1=5$

$a_2=-4,b_2=2,c_2=-k$

Now,

$\dfrac{-2}{-4}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{5}{-k}$

On cross multiplication, we get

$-k=10$

$k=-10$

Therefore, the missing value is -10.

6 0

3 years ago