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sveticcg [70]
3 years ago
8

Please help me with 11-16 please! Thank you

Mathematics
1 answer:
sasho [114]3 years ago
3 0

11. -1

12. -9

13. 32

14. -15

15. 10

16. 24

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Triangle HIJ has coordinates H(-3,5), (8,2), and J(-3,-5). Find
Vlad1618 [11]

Answer:

H (x= -3 , y=5)

I (x=8 , y=2)

J (x= -3 , y= -5)

D is the distance between the 2 points so just plug in the x and y cords of each of the corresponding points

so for (HI)

you plug in 2 on the red x

and 8 on the red y

then -3 in the blue x

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Multiply -3/10xy(60xy^6)
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= -18x^2y^7

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3 years ago
Identify an equation in slope – intercept‘s form for the line parallel to Y=5x+2 that passes through (-6,-1).
maksim [4K]

Slope-intercept form of a line is y=mx+b where m= slope and b=y-intercept.

First step is to compare this line with the given line y=5x+2 to get the value of m.

After comparing we will get m=5.

Now slope of all parallel lines are equal which means if slope of this line is 5 then slope of the line which is parallel to this line will also be 5.

Point- slope form of a line is:

y-y_{1} =m(x-x_{1} )

Given the line passes through (-6,-1). So, plug in x1=-6, y1=-1 and m=5 in the above equation. So,

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y+1=5(x+6)

y+1=5x+30

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8 0
3 years ago
Which comparison is not correct?<br> 06&gt;5<br> 0-7&gt;-9<br> 0-3&lt;4<br> O 4 &lt;-12
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#4

Step-by-step explanation:

7 0
3 years ago
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The cross-sectional areas of a right pyramid and a right cylinder are congruent. The right pyramid has a height of 10 units, and
Ilia_Sergeevich [38]

Answer:

The volume of the cylinder is 2.1 times the volume of the pyramid

Step-by-step explanation:

step 1

Find the volume of the pyramid

we know that

The volume of a right pyramid is equal to

V=\frac{1}{3}BH

where

B is the area of the base of pyramid

H is the height of the pyramid

we have

H=10\ units

substitute

Vp=\frac{10}{3}B\ units^{3}

step 2

Find the volume of the right cylinder

we know that

The volume of right cylinder is equal to

V=BH

where

B is the area of the base of cylinder

H is the height of the cylinder

we have

H=7\ units

substitute

Vc=7B\ units^{3}

step 3

Compare the volumes

we know that

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Find the ratio of their volumes

\frac{Vp}{Vc}

substitute

\frac{Vp}{Vc}=\frac{(\frac{10}{3}B)}{7B}

Simplify

\frac{Vp}{Vc}=\frac{10}{21}

21Vp=10Vc\\\\Vc=2.1Vp

therefore

The volume of the cylinder is 2.1 times the volume of the pyramid

7 0
3 years ago
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