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3 years ago

What is 15/2 divided by 25/8.

1 answer:

7 0

$\frac{15}{2} \times \frac{8}{25 } = \frac{15}{2} \div \frac{25}{8}$
and 15×8=120 and 2×25= 50
so the answer is
$\frac{120}{50}$
simplifies to
$2 \frac{2}{5}$

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Can a -150/10 be classified as both a rational number and a integer

irakobra [83]

Yes! This is because -150/10 can be simplified to be -15, which is a rational number.

The word “rational” sounds like another math word you’ve heard of before. Do you know what it is?

Well, it’s “ratio”!! Ratios can be seen in the forms x:y and x/y.

ANY RATIONAL NUMBER HAS THE ABILITY TO BE WRITTEN AS A RATIO!! This will completely exclude numbers with super long decimal points (ex: 1.2345678809928374737272828...)

This number also meets the requirements of being an integer. An integer is any whole number (this excludes decimals and fractions)

I know it’s written as a fraction. However, the fraction could be simplified, making it -15, which means this is both a rational number and an integer!!

6 0

3 years ago

The value of -9(square root) is not-3 because

love history [14]

Answer:

A) $(-3)^{2} \neq -9$

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

I NEED HELP!!<br>answer from the image below <br><br>m∠BCD =

nevsk [136]

Answer:

28 degrees

Step-by-step explanation:

If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.The corresponding angles are the ones at the same location at each intersection.

In the figure ABD and EDF are corresponding angles. So they are equal

So equation the angle ABD and EDF , we get

(3x+4) = (7x-20)

Group the like terms,

3x-7x = -20-4

-4x = -24

$x = \frac{-24}{-4}$

x = 6

Thus BCD will be,

(6x - 8)

=>(6(6)-8)

=>(36-8)

=> 28 degrees

4 0

3 years ago

Solve ln(6x-1) = -4 for x

Reika [66]

Answer: Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form:

−

Decimal Form:

−

0.5

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

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

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

2 years ago