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krok68 [10]
2 years ago
15

Can Someone help me ​

Mathematics
2 answers:
tresset_1 [31]2 years ago
8 0

Answer:

it's c

Step-by-step explanation:

it's c because I know math lol

Westkost [7]2 years ago
3 0

Answer:

c.

Step-by-step explanation:

because it says monthly fix cost and mothly means x

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How long is crosspiece AB? (Give answer to tenths accuracy.)
GalinKa [24]
a^{2}  + b^{2}=  c^{2}
4^{2} +10^{2}=c^{2}
16+100=116
c= [/tex]116
c= 10.77
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3 years ago
Plz help, it’s been due for a long time now but I need help.
natima [27]

Answer:

< 4, < 10 corresponding

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< 8, < 12 other (opposite exterior)

< 6, < 12 alternate interior

< 1, < 2 other (supplementary)

< 15, < 11 alternate interior

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Two lines that are stretched into infinity and still never intersect are called co-planar lines and are said to be parallel lines. ... Corresponding angles are congruent if the two lines are parallel. All angles that have the same position with regards to the parallel lines and the transversal are corresponding pairs.

6 0
2 years ago
If f(x) = 3x^2 - x, find f(-2).
saw5 [17]

Answer:

14

Step-by-step explanation:

to find f(value), we plug that x value into the equation

so, if..

f(x)= 3x² -x

f(-2)= 3 · (-2)² - (-2)

f (-2) = 3 · 4 + 2

f (-2) = 12 + 2

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4 0
1 year ago
Read 2 more answers
Find the equation of a line passing through (3, -5) that is parallel to 2x + 6y =10
PtichkaEL [24]

Answer:

The equation of the line would be y = -1/3x - 4

Step-by-step explanation:

In order to find this, we first need to find the slope of the original line. We do this by solving for y.

2x + 6y = 10

6y = -2x + 10

y = -1/3x + 5/3

Now that we see the slope as -1/3, we know the new line will have the same slope thanks to the definition of parallel lines. So, we can use this slope and the point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - -5 = -1/3(x - 3)

y + 5 = -1/3x + 1

y = -1/3x - 4

7 0
3 years ago
Help me pls !!!!!!!!!!
Irina18 [472]

Answer: What do u need help with?

Step-by-step explanation:

6 0
2 years ago
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