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3 years ago

I need help!!! what is the highest common factor of 30 and 60?

Mathematics

2 answers:

PtichkaEL [24]3 years ago

8 0

<span>The Greates Common Factor (GCF) is: 2 x 3 x 5 = 30</span>

Agata [3.3K]3 years ago

7 0

I thought it was 10 because 10 x 3 and 10 x6 but it is 2

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Rectangle B is a scale drawing of rectangle A using a scale factor of 150%. The area of rectangle A is 32 square units. What is

drek231 [11]

Answer:

The answer is 72 square units

Step-by-step explanation:

If rectangle A is scaled by 150% that means the length and breadth of rectangle A were both increased by 1.5(150%) to yield rectangle B.

So, the area of rectangle B is 1.5L x 1.5B = 2.25LB

But, LB = area of rectangle A which is = 32

So, area of rectangle B

= 2.25 x 32

= 72square units.

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2 years ago

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Please answer this math equation.

nasty-shy [4]

The interval where the function is nonlinear and decreasing is 0 < x < 4

<h3>How to determine the interval where the function is nonlinear and decreasing?</h3>

The straight lines on the graph are the intervals where the graph is linear

This means that the straight lines on the graph will not be considered

Considering the curve, the graph decrease from x = 0 to x = 4

This can be rewritten as:

0 < x < 4

Hence, the interval where the function is nonlinear and decreasing is 0 < x < 4

Read more about function intervals at:

brainly.com/question/13136492

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2 years ago

I need to know how to solver this coz I needy grades up

Ne4ueva [31]

Umm y = 2× + 8 ? Hard

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3 years ago

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What is the exponential regression equation for the data set?

Ne4ueva [31]

Answer:

Exponential Regression. An exponential regression is the process of finding the equation of the exponential function that fits best for a set of data. As a result, we get an equation of the form y=abx where a≠0 .

Step-by-step explanation:

6 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago