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LiRa [457]
3 years ago
9

What is the equation of its axis symmetry?

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
6 0

Answer:

y=0

Step-by-step explanation:

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Which of these has the largest value?<br><br>A 1¹<br>B 2¹<br>C 2³<br>D 3²<br>​
Anettt [7]
D has the largest value
6 0
2 years ago
Add in write in the simplest form 3 1/3 divided by 2 3/5
bixtya [17]

Answer:

1 11/39

Step-by-step explanation:

3 1/3 = 10/3

2 3/5 = 13/5

10/3 / 13/5

50/39 = 1 11/39

BRAINLIEST WOULD MEAN A TON ;)

5 0
3 years ago
If EF bisects angle CEB, angle CEF=7x + 21 and angle FEB = 10x - 3, find the measure of DEB.
morpeh [17]

Answer: DEb = 26°

Step-by-step explanation:

<u>Given information</u>

CEF = 7x + 21

FEB = 10x - 3

<u>Given expression deducted from the definition of the bisector</u>

FEB = CEF

<u>Substitute values into the expression</u>

10x - 3 = 7x + 21

<u>Subtract 7x on both sides</u>

10x - 3 - 7x = 7x + 21 - 7x

3x - 3 = 21

<u>Add 3 on both sides</u>

3x - 3 + 3 = 21 + 3

3x = 24

<u>Divide 3 on both sides</u>

3x / 3 = 24 / 3

x = 8

<u>Find the sum of the angle of CEF and FEB</u>

 7x + 21 + 10x - 3

=7 (8) + 21 + 10 (8) - 3

=56 + 21 + 80 - 3

=77 + 80 - 3

=157 - 3

=154

<u>Subtract 154 from the straight angle</u>

DEB = 180 - 154

\boxed{DEB = 26}

Hope this helps!! :)

Please let me know if you have any questions

8 0
2 years ago
Solve for X<br><br> Please help
Nookie1986 [14]

Answer:

x = 5

Step-by-step explanation:

set JN and NL equal to each other...

14x – 17 = 3x + 38

11x = 55

x = 5

5 0
2 years ago
A force of 12 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i
sattari [20]

Answer:

Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:

Normal length of the spring = 8 in or \frac{8}{12} ft

= \frac{2}{3} ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= \frac{11}{12} ft

Force applied to stretch the spring = 12 lb

By Hook's law,

F = kx [where k is the spring constant and x = length by which the spring is stretched]

12 = k(\frac{2}{3})

k = 18

Work done (W) to stretch the spring by \frac{11}{12} ft will be

W = \int\limits^\frac{11}{12} _0 {kx} \, dx

    = \int\limits^\frac{11}{12} _0 {(18x)} \, dx

    = 18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0

    = 9(\frac{11}{12})²

    = 7.56 lb-ft

6 0
3 years ago
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