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Andreyy89
3 years ago
10

Which function represents the graph

Mathematics
1 answer:
natulia [17]3 years ago
4 0

Answer:

A

Step-by-step explanation:

the function is moved up by 8 on the y axis which means you need a +8 that isn't directly attached to the x

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Take a factor out of the square root:<br> sqrt(3y^2) , where y&lt;0
elena-s [515]

factor took out was y^2 , And resultant expression is  y\sqrt{3} .

<u>Step-by-step explanation:</u>

Here we need to Take a factor out of the square root:  sqrt(3y^2) , where y<0 .Let's do this :

The given expression is  sqrt(3y^2) or , \sqrt{(3y^2)}

We know that , to take out a factor out of the square root that factor must have a degree in multiple of 2 , as   x^2,x^4,x^16,(x+6)^2,(x-2)^4  etc . So ,

⇒ \sqrt{3y^2}

⇒ \sqrt{3}(\sqrt{y^2})

⇒ \sqrt{3}(y^2)^{\frac{1}{2}

⇒ \sqrt{3}(y)^{\frac{2}{2}

⇒ y\sqrt{3}

Therefore , factor took out was y^2 , And resultant expression is  y\sqrt{3} .

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3 years ago
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EastWind [94]
El resultado de 79 mas X igual 452 es 141
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3 years ago
How would you find the surface area of the figure represented by the given net? (Note: The area of each circular base is the sam
ioda

Answer:

Add 1,105.28 + 200.96 + 200.96.

Step-by-step explanation:

we know that

The surface area of the cylinder is equal to

SA=2B+LA

where

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LA is the lateral area

In this problem we have

B=200.96\ mm^{2}

LA=1,105.28\ mm^{2}

substitute

SA=2(200.96)+1,105.28

SA=1,105.28+200.96+200.96

SA=1,507.2\ mm^{2}

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4 years ago
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Answer:

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Step-by-step explanation:

Let A(a , b) be a point of the line

and m be the slope.

The equation of the line in Point-slope form :

y - b = m (x - a).

…………………………

Given :

Slope = -2

A(4 , -6)

Then

Point-slope equation : y + 6= -2(x - 4)

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2 years ago
Geometry question help please!
valentinak56 [21]
UW; Converse of the Isosceles Triangle Theorem

This is the answer because angles T and W are congruent. Meaning that they make an isosceles triangle. The two sides connecting to those angles should be congruent. Therefore UT and UW are congruent. 
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