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3 years ago

4√45·2√12? Help ASAP how do I do this will mark brainliest

Mathematics

1 answer:

Paraphin [41]3 years ago

6 0

Multiply the numbers outside the square root together and the two square roots together to get a single term and simplify

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What number is 15% of 60

zhannawk [14.2K]

0.15 * 60 = 9 or 15/100 * 60/1 = 900/100 = 9

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3 years ago

Read 2 more answers

I AM STUCK!!! PLEASE HELP!!!!!

mixer [17]

A = 0.1
B = 50.9

These are rounded to the nearest tenth

8 0

3 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

5 0

2 years ago

The sales tax in one town is 8%. So, the total cost of an item can be written as c+0.08c. What is the total cost of an item that

Nezavi [6.7K]

Try this: 1.08($12) = $12.96 (total cost)

7 0

2 years ago

Does anyone know how to do these

creativ13 [48]

Answer:

Step-by-step explanation:\

my quick answer for you would be to look is there is a website on the paper so that way you can search online and see what they give you.

5 0

2 years ago