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3 years ago

Find the missing numerator in each of the following fractiions? A. 5/6= /12. B. 2/4= /36. C. 2/5=/20. D. 4/25=/100

Mathematics

1 answer:

lesantik [10]3 years ago

5 0

<u>Answer:</u>

A. 10

B. 18

C. 8

D. 16

<u>Step-by-step explanation:</u>

We are given a number of incomplete fractions and we are to find the missing numerator in each one of these.

To find the missing numerator or denominator, we multiply the numerator and the denominator by the same number to get an equivalent fraction.

A. $\frac{5}{6} =\frac{10}{12}$

Here the denominator is multiplied by 2 so the denominator will also be multiplied by 2 to get 10.

B. $\frac{2}{4} =\frac{18}{36}$

Numerator is multiplied by 9.

C. $\frac{2}{5} =\frac{8}{20}$

Numerator is multiplied by 4.

D. $\frac{4}{25} =\frac{16}{100}$

Numerator is multiplied by 4.

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One number is 5 more than another number. their sum is −15 . what are the numbers?

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X=5+y
x+y=-15

subsitute 5+y for x

5+y+y=-15
5+2y=-15
minus 5 from both sides
2y=-20
divide both sides by 2
y=-10

sub back
x=5+y
x=5-10
x=-5

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Step-by-step explanation:

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hakim wrote an equation to represent the number of meters he would run for each kilometer, k, in a long distance race.W hat is t

ASHA 777 [7]

K=500 because of the kilometer

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2 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

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2 years ago

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topjm [15]

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3 years ago