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Nina [5.8K]
3 years ago
10

What is the missing side

Mathematics
1 answer:
igomit [66]3 years ago
8 0
X = 5 inches the answer is 
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Which is the solution to the inequality |x-4|<3
pishuonlain [190]

Answer:

\large\boxed{1

Step-by-step explanation:

|x-4|

8 0
3 years ago
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What is the value of x in the product of powers below?<br><br> help me please
belka [17]
6^9 x 6^-7 = 6^2

x= -7

Hope this helps :)
3 0
3 years ago
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help with number 2 please i have no idea how to do this. it would be very helpful if you can show the steps
zloy xaker [14]
So in this case, we have to replace the known value.
y=3
y=-2x+3

3=-2x+3

Then we leave our unknown value alone.
\frac{3-3}{2}= x

In this case, our x value would be 0.

We check it...
3=-2(0)+3
3=0+3
3=3

So y=3 x=0

For the second one we have...
y=3x+2
y=-3x-4

For this we substitute the y in any of the equation...
3x+2=-3x-4

We move the unknown values to one side and the ones without unlown values to the other side...
3x+3x=-4-2

Then we solve
6x=-6

Then we leave the unknown value alone.
x=\frac{-6}{6}

Then solve for x.
x= -1

Then for our y value we return to one of the original equations and substitute the x value.
y=3x+2
y=3(-1)+2
y=-3+2
y=-1

y=-3x-4
y=-3(-1)-4
y=3-4
y=-1

So in this case we got that x= -1 and y= -1
4 0
3 years ago
if you repeat the perpendicular line segment construction twice using paper folding, you can construct:
ehidna [41]

Answer:

The correct answer is  option C.

The mid point of the line segment.

Step-by-step explanation:

the perpendicular line segment construction twice using paper folding

we have to find the mid point of the given line segment.

We get the midpoint easily when fold the paper correctly

Therefore the correct answer is  option C.

The mid point of the line segment.

6 0
3 years ago
Read 2 more answers
Hurry please!!! Fast response
AlladinOne [14]

Answer:

A.

SA= \frac{2}{3} \pi rl+16 \pi r^2

Step-by-step explanation:

Surface area of the original cone

SA= \pi r l + \pi r^2... (1)

When, radius is quadrupled and slant height is reduced to one sixth

i. e. \: \:r = 4r\: \: \&\:\: l= \frac{1}{6}l

Plug the above values of r and l in equation (1), new surface area becomes:

SA= \pi (4r)\times \frac{1}{6} l+ \pi (4r)^2

SA= \frac{4}{6} \pi rl+16 \pi r^2

\huge \purple {\boxed {SA= \frac{2}{3} \pi rl+16 \pi r^2}}

8 0
2 years ago
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