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k0ka [10]
3 years ago
5

Help me solve this problem please

Mathematics
1 answer:
ASHA 777 [7]3 years ago
5 0

Answer:

3x plus 5 is definetely greater than -16 cause of the negative and 8 is bigger also then a negative

Step-by-step explanation:

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−8/9 ⋅ 5/6<br> What is the product in simplest form?<br><br><br> Fractions btw
klio [65]
The answer 56 it’s already put in simplest form
7 0
3 years ago
(-3x^2 +11x -1)+(7x^2 - 4x + 4)
muminat

Answer: what answer do you want?

Step-by-step explanation:

4 0
3 years ago
The missing term in the denominator is
Margaret [11]

\frac{7^{16} }{7^{12} }= \frac{7^{16-34} }{7^{12-34} } =\frac{7^{-18} }{7^{-22} } = 2401

so the missing term in the denominator is 7^(-22)

ok done. Thank to me :>

5 0
3 years ago
PLEASE HELP W/ THIS ASAP!!
Leni [432]

Answer:

- 1.375

Step-by-step explanation:

Each dot is worth -1/8 of a unit when moving left.

So the coordinates of the dark dot are - 1 and - 3/8

or - 1.375 . The word coordinates (plural) bothers me. It all depends on how you have been asked to represent these points. Another possibility is (-1.375,0). You have to choose from the examples you have been given. I wouldn't do that because you have not been given and xy graph.

3 0
3 years ago
The midpoint of AB is M (6,1). If the coordinates of A are (4, 8), what are the coordinates of B?
geniusboy [140]

Answer:

B = (8,-6)

Step-by-step explanation:

Given

A = (4,8)

M = (6,1) -- Midpoint

Required

Find B

This question will be solved using midpoint formula;

M(x,y) = (\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})

Where

A(x_1,y_1) = (4,8)

M(x,y) = (6,1)

Substitute these values in the midpoint formula;

(6,1) = (\frac{4 + x_2}{2},\frac{8+y_2}{2})

By comparison; we have:

\frac{4 + x_2}{2} = 6 -- (1)

\frac{8 + y_2}{2} = 1 -- (2)

Solving (1)

\frac{4 + x_2}{2} = 6

Multiply both sides by 2

4 + x_2 = 12

Subtract 4 from both sides

x_2 = 8

Solving (2)

\frac{8 + y_2}{2} = 1

Multiply both sides by 2

8 + y_2 = 2

Subtract 8 from both sides

y_2 = -6

Hence:

B = (8,-6)

3 0
3 years ago
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