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3 years ago

jennifer scored 75 on her quiz last week. this week she scored 90. what was the percent of increase on the quiz score?

Mathematics

1 answer:

Arlecino [84]3 years ago

4 0

Answer:

The answer is a 20% increase

Hope this helps!

Mark me brainliest if I'm right :)

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Find the sum of the first 6 terms of the following series, to the nearest

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Answer:277

Step-by-step explanation:

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3 years ago

Mr. Chen started doing yard work at 10:00. He watered flowers for 6 minutes, weeded his garden for 12 minutes, and trimmed bushe

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Step-by-step explanation:idk

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3 years ago

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Consider the equality xy k. Write the following inverse proportion: y is inversely proportional to x. When y = 12, x=5.

skelet666 [1.2K]

Answer:

$y=\dfrac {60} {x}$ or $xy=60$ (depending on your teacher's format preference)

Step-by-step explanation:

<h3><u>Proportionality background</u></h3>

Proportionality is sometimes called "variation". (ex. " 'y' varies inversely as 'x' ")

There are two main types of proportionality/variation:

Direct
Inverse.

Every proportionality, regardless of whether it is direct or inverse, will have a constant of proportionality (I'm going to call it "k").

Below are several different examples of both types of proportionality, and how they might be stated in words:

$y=kx$ y is directly proportional to x
$y=kx^2$ y is directly proportional to x squared
$y=kx^3$ y is directly proportional to x cubed
$y=k\sqrt{x}}$ y is directly proportional to the square root of x
$y=\dfrac {k} {x}$ y is inversely proportional to x
$y=\dfrac {k} {x^2}$ y is inversely proportional to x squared

From these examples, we see that two things:

things that are <u>directly proportional</u> -- the thing is <u>multipli</u>ed to the constant of proportionality "k"
things that are <u>inversely proportional</u> -- the thing is <u>divide</u>d from the constant of proportionality "k".

<h3><u>Looking at our question</u></h3>

In our question, y is inversely proportional to x, so the equation we're looking at is the following $y=\dfrac {k} {x}$ .

It isn't yet clear what the constant of proportionality "k" is for this situation, but we are given enough information to solve for it: "When y=12, x=5."

We can substitute this known relationship pair, and find the "k" that relates this pair of numbers:

<h3><u>Solving for k, and finding the general equation</u></h3>

General Inverse variation equation...

$y=\dfrac {k} {x}$

Substituting known values...

$(12)=\dfrac {k} {(5)}$

Multiplying both sides by 5...

$(12)*5= \left ( \dfrac {k} {5} \right ) *5$

Simplifying/arithmetic...

$60=k$

So, for our situation, k=60. So the inverse proportionality relationship equation for this situation is $y=\dfrac {60} {x}$ .

The way your question is phrased, they may prefer the form: $xy=60$

7 0

2 years ago

Please help!!! Verify which of the following are identities.

Ierofanga [76]

Answer:

First one

Step-by-step explanation:

None of the equations are identities

4 0

3 years ago

Factorise x^2/4-y^2/4

Nuetrik [128]

Answer:

$\large\boxed{\left(\dfrac{x}{2}-\dfrac{y}{2}\right)\left(\dfrac{x}{2}+\dfrac{y}{2}\right)=\dfrac{x-y}{2}\cdot\dfrac{x+y}{2}}$

Step-by-step explanation:

$\dfrac{x^2}{4}-\dfrac{y^2}{4}=\dfrac{x^2}{2^2}-\dfrac{y^2}{2^2}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\=\left(\dfrac{x}{2}\right)^2-\left(\dfrac{y}{2}\right)^2\qquad\text{use}\ (a-b)(a+b)=a^2-b^2\\\\=\left(\dfrac{x}{2}-\dfrac{y}{2}\right)\left(\dfrac{x}{2}+\dfrac{y}{2}\right)=\dfrac{x-y}{2}\cdot\dfrac{x+y}{2}$

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3 years ago