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2 years ago

4:X is equivalent to 24:12 X=???

Mathematics

2 answers:

Elena-2011 [213]2 years ago

6 0

Answer:

X=2

Step-by-step explanation:

Ratios

4:x = 24:12

24:12 reduced is 2:1

Therefore we can see x must be half of 4.

So the answer is 2

Hope this helps! :)

pls mark brainiest

nasty-shy [4]2 years ago

6 0

Answer:

X = 2

Step-by-step explanation:

4:X is equivalent to 24:12

4 is multiplied by 6 to get to 24 so divide each term by 6

24/6:12/6

4 : 2

X = 2

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AnnyKZ [126]

Answer:

50 pens

Step-by-step explanation:

x= 25 percent of 200

25/100 times X/200

Cross multiply them and you get

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Which means they have away 50 pens

7 0

2 years ago

Read 2 more answers

I have this point: (-1,2) and my slope is undefined. How would I write my equation in point-slope form: y - y^1 = m (x -x^1)? Th

tiny-mole [99]

Answer:

x = -1

Step-by-step explanation:

Given the point, (-1, 2), and that the slope is undefined.

The standard linear equation of vertical lines is x = a, where the x-intercept is (a, 0), and the slope is undefined because all points on the line have the same x-coordinate. Attempting to solve for the slope of a vertical line using the slope formula, m = (y₂ - y₁)/(x₂ - x₁), will result in a mathematical operation of division by zero (which is an undefined operation).

Since the slope is undefined, then it is not possible to create a linear equation in either the slope-intercept form, or point-slope form.

Therefore, the equation of a vertical line given the point, (-1, 2) is x = -1.

7 0

2 years ago

NEED HELP BEING TIMED

Virty [35]

Answer:

The answer is A.

The value of the 1 in the one's place is 10 times the value of the 1 in the tenths place.

Step-by-step explanation:

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2 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$