All categories

3 years ago

Which number is equal to 100?

Mathematics

2 answers:

Tcecarenko [31]3 years ago

6 0

Answer:

10 square2

36 + 64 = 100. In other words, 6² + 8² = 10²

VladimirAG [237]3 years ago

4 0

Answer:

C. 10^2

Step-by-step explanation:

When you have an exponent and the base is 10, the exponent is how many zeros there are total that come after the 1. There are 2 zeros after the 1 in 100, so 10^2 is the answer.

You might be interested in

What number should you subtract from the numerator of the fraction 52367/47633 and add to the denominator so that, after reducin

nata0808 [166]

Let $x$ be the number that you should subtract from the numerator of the fraction 52367/47633 and add to the denominator so that, after reducing, you get 17/38.
 $\frac{52367-x}{47633+x} = \frac{17}{38}$
$38(52367-x)=17(47633+x)$
$1989946-38x=809761+17x$
$1180185=55x$
$x= \frac{1180185}{55}$
$x= \frac{236037}{11}$

We can conclude that the number you should subtract from the numerator of the fraction 52367/47633 and add to the denominator so that, after reducing, you get 17/38 is: $\frac{236037}{11}$

4 0

3 years ago

What is equivalent to 2:6

Katena32 [7]

HI,THE ANSWER TO YOUR PROBLEM IS 1:3

HOPE THIS HELPED! D:

8 0

4 years ago

Read 2 more answers

What is the value of the function at x = –2? 0 –1 –6 2

makvit [3.9K]

I’m pretty sure it’s 0

6 0

3 years ago

Jackie Kessel has found a home she would like to purchase.The seller accepts her offer of $210,900. Jackie will make a down paym

Katen [24]

Answer:

$179,265

Step-by-step explanation:

Subtract the 15% down payment from the purchase price ($210,900) to obtain the amount still due, for which Jackie will apply for a mortgage. Alternatively, multiply $210,900 by 0.85, obtaining $179,265. This would be the mortgage amount.

4 0

3 years ago

Find the quotient of 21x^2y^6+6xy^3-30xy/3xy.

kaheart [24]

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------$

$\bf \cfrac{21x^2y^6+6xy^3-30xy}{3xy}\implies \stackrel{\textit{distributing the denominator}}{\cfrac{21x^2y^6}{3xy}+\cfrac{6xy^3}{3xy}-\cfrac{30xy}{3xy}}
\\\\\\
\cfrac{21}{3}\cdot x^2x^{-1}y^6y^{-1}+\cfrac{6}{3}\cdot x^1x^{-1}y^3y^{-1}-\cfrac{30}{3}\cdot x^1x^{-1}y^1y^{-1}
\\\\\\
7x^{2-1}y^{6-1}+2x^{1-1}y^{3-1}-10x^{1-1}y^{1-1}\implies 7xy^5+2y^2-10$

3 0

3 years ago

Read 2 more answers