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exis [7]
3 years ago
12

PLEASE HELPPPPP ( will make brainliest )

Mathematics
1 answer:
frutty [35]3 years ago
6 0
I believe you are right, it is C

If you go through each option, it would be correct.

-5(7) = -35 + 3 = -32
-3(7) = -21 + 3 = -18
-1(7) = -7 + 3 = -4
6(7) = 42 + 3 = 45
7(7) = 49 + 3 = 52

So the equation y= 7x+3, as you selected would be correct.

Please correct me if I am wrong and have a good day!
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**The given angles are (-1+38x) and (36x+3)".<br><br> x=
topjm [15]

Answer:

x = 2

Step-by-step explanation:

these angles are alternate-interior angles which are congruent

-1 + 38x = 36x + 3

38x = 36x + 4

2x = 4

x = 2

7 0
3 years ago
Which angles are right? Check all that apply
Hitman42 [59]

Answer:

If I remember correctly, right angles are 90 degree angles. So the answers would be A. B. and C.

Step-by-step explanation:

If you see the little box on the angle, that means it's a 90 degree (right) angle.

5 0
3 years ago
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Factor 14x2+6x-7x-3 by grouping
QveST [7]

14x^2+6x-7x-3

14x^-7x  +6x-3

group

7x(2x-1) +3(2x-1)

factor out (2x-1)


(2x-1)(7x+3)

5 0
3 years ago
A store sells a 33-pound bag of oranges for \$ 3.60$3.60 and a 55-pound bag of oranges for \$ 5.25$5.25. What is the difference
katrin [286]

Answer:

0.01364

Step-by-step explanation:

It is given that,

A store sells a 33-pound bag of oranges for $3.60 and a 55-pound bag of oranges for $5.25.

Price per pound of 33 pound bag is 3.60/33 = 0.10909 price per pound

Price per pound of 55 pound bag of oranges is 5.25/55 = 0.09545 price per pound

Difference between price per pound for the 33-pound bag of oranges and the price per pound for the 55-pound bag of oranges is :

D = 0.10909 - 0.09545

D = 0.01364

Therefore, this is the required solution.

3 0
3 years ago
Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.
Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If x + \dfrac{1}{x} is an integer

∴ \left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

x + \dfrac{1}{x}

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}

By simplification of the cube of the given integer expressions, we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )

Therefore, we have;

\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}

By rearranging, we get;

x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )

Given that  x + \dfrac{1}{x} is an integer, from the closure property, the product of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 is an integer and 3\cdot \left (x + \dfrac{1}{x} \right ) is also an integer

Similarly the sum of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

\therefore x^3 + \dfrac{1}{x^3} =   \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer.

4 0
3 years ago
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