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3 years ago

If 75% of a number is 60, what is 20% of the number

Mathematics

2 answers:

Colt1911 [192]3 years ago

6 0

I hope this is right i think the answer is 40

Elan Coil [88]3 years ago

3 0

I think it is 16 how i got that is by finding the number that 60 was 75% of which is 80 then i took 80 * 20 / 100 to get 16 which should 20 precent of 80 hope that helps

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Find x, y, and z.

earnstyle [38]

Answer:

x=4.5,y=5,z=9

Step-by-step explanation:

First, we use the pythagorean theorem in the lower right triangle (with sides 4,x,6) and find x^2+4^2=6^2 and therefore x^2=6^2-4^2=20 so x=√20 which is approximately 4.5.

Also, we can see from the picture that z=y+4, and therefore the only option left is x=4.5,y=5,z=9.

5 0

2 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

In art class Ms smith is working on polygons. She want the students to Create a picture of the polygons. She points out to the c

Ivan

Answer:

There are two more sides on a hexagon than on a quadrilateral

Step-by-step explanation:

If the hexagon has 6 sides, and the quadrilateral has 4, then 6-4=2

8 0

3 years ago

PHYSICS As a ray of light meets a mirror, the light

Lady bird [3.3K]

<h3>Answer: 37 degrees</h3>

========================================================

Explanation:

The angle of incidence is equal to the angle of reflection. The angle of light coming in is the same as the angle of light bouncing out. Each angle is made with the dashed line as the diagram shows.

So angle RMI = 106 is bisected, i.e. divided in half, to get

Angle RMN = 53
Angle NMI = 53

since 106/2 = 53

Then notice that the horizontal dashed line is perpendicular to the vertical mirror. This means angle NMJ is a 90 degree angle and,

angle RMJ = (angle NMJ) - (angle RMN)

angle RMJ = 90 - 53

angle RMJ = 37 degrees

----------

In short: divide 106 in half to get 53, then subtract it from 90

5 0

1 year ago

What is 38,800 rounded to the nearest hundred mark all that apply?

SashulF [63]

38,800 is the answer

4 0

3 years ago