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jarptica [38.1K]
3 years ago
9

Please need help fast. Look at the picture. Please explain.

Mathematics
1 answer:
Zigmanuir [339]3 years ago
7 0
The answer is 17

This is because 1 + 2 must equal 180 as they are in a straight line.
So 4x + 2 + 110 = 180
Take away 110 from both sides
4x + 2 = 70
Take away 2 from both sides to get 4x = 68. Finally, divide both sides by 4 to get x = 17
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I NEED HELP ASAP PLEASE PLEASE HELP ME
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Answer:

36 + 42 = 6(6 + 72)

Step-by-step explanation:

36 + 42 = 78 - 6 = 72 = 6(6 + 72)

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Two-step equations - integers
Andrews [41]
Question 1. -3 + v5 = 0
Work : -3 + 5v = 0
5v = 3 ( divide )
answer v = 3/5

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work : -7b = -20 - 1
-7b = -21 ( divide )
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3 0
3 years ago
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An internet service provider is implementing a new program based on the number of connected devices in each household. Currently
oee [108]
Cost of the current plan: $175
Number of devices: x
Cost of the new plan: $94+($4.50/device)x

Cost of the new plan is less than the current plan:
$94+($4.50/device)x<$175

Solving for x:
$94+($4.50/device)x-$94<$175-$94
($4.50/device)x<$81
(device/$4.50)($4.50/device)x<(device/$4.50)($81)
x<18 devices

Please, see the attached file.
Thanks.

8 0
3 years ago
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The equation of line AG is y=_b_x. The midpoint of
Lina20 [59]

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D

Step-by-step explanation:

7 0
3 years ago
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Which expressions are equivalent to \dfrac{4^{-3}}{4^{-1}} 4 −1 4 −3 ​ start fraction, 4, start superscript, minus, 3, end super
Slav-nsk [51]

Answer:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}

\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}

Step-by-step explanation:

Given

\dfrac{4^{-3}}{4^{-1}}

Required

Choose equivalent expressions

Choosing the first answer:

\dfrac{4^{-3}}{4^{-1}}

Split expressions

4^{-3} * \frac{1}{4^{-1}}

Apply laws of indices: (a^{-b} = \frac{1}{a^b})

\frac{1}{4^3} * \frac{1}{4^{-1}}

Apply laws of indices: (a^{-b} = \frac{1}{a^b})

\frac{1}{4^3} * \frac{1}{1/4}

\frac{1}{4^3} * \frac{4^1}{1}

\frac{4^1}{4^3}

Hence:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{4^{1}}{4^{3}}

Choosing the second:

\dfrac{4^{-3}}{4^{-1}}

Apply law of indices: (\frac{a^m}{a^n} = a^{m-n})

So,

\dfrac{4^{-3}}{4^{-1}} = 4^{-3-(-1)}

\dfrac{4^{-3}}{4^{-1}} = 4^{-3+1)}

\dfrac{4^{-3}}{4^{-1}} = 4^{-2}

Apply law of indices: (a^{-b} = \frac{1}{a^b})

So:

\dfrac{4^{-3}}{4^{-1}} = \dfrac{1}{4^{2}}

4 0
3 years ago
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