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serious [3.7K]
2 years ago
12

Solve this system of linear equations. Separate

Mathematics
1 answer:
Yakvenalex [24]2 years ago
6 0

Answer:

5x

Step-by-step explanation:

im not to sure tbh

You might be interested in
What is the solution to y + 7 = -9 A. y = -16 B. y = -2 C. y = 2 D. y = 16
Helen [10]

Answer:

A. y = -16

Step-by-step explanation:

Hello!

What we do to one side of the equation we have to do to the other.

y + 7 = -9

We have to get y by itself so we have to get rid of the seven by doing the opposite of what it says

The opposite of addition is subtraction so we subtract 7 from both sides

y + 7 - 7 = -9 - 7

Solve

y + 0 = -16

Simplify

y = -16

The answer is A. y = -16

Hope this helps!

4 0
3 years ago
I need to know the answer to p=4x-140
dalvyx [7]
This is not look like something that could be answered, check around the paper for a graphing box.
This looks like a formula for a line on a graph: y=Mx+b. I can help you if this is the case, 1. Put a point at -140 on the Y axis (up and down) 2. Move up 1 and over 4and put a dot there( you could multiply the 1 and 4 to cover a larger area) because it goes all the way to -140.
7 0
2 years ago
Whats the answer giving brainliest>>>>
Elza [17]

Answer:

the answer is b

Step-by-step explanation:

brainliest pls

5 0
3 years ago
Write a polynomial function of least degree with the given zero. -1+ √ 2, √ 3
kirill115 [55]

Answer:

f(x) = x^{4} + 2x³ - 4x² - 6x + 3

Step-by-step explanation:

Note that radical zeros occur in conjugate pairs, thus

- 1 + \sqrt{2} is a zero then - 1 - \sqrt{2} is also a zero

\sqrt{3} is a zero then - \sqrt{3} is also a zero

Thus the corresponding factors are

(x - (- 1 + \sqrt{2}) ), (x - (- 1 - \sqrt{2}) ), (x - \sqrt{3}), (x - (- \sqrt{3})), that is

(x + 1 - \sqrt{2}), (x + 1 + \sqrt{2}), (x - \sqrt{3}), (x + \sqrt{3})

The polynomial is then the product of the roots

f(x) = (x + 1 - \sqrt{2})(x + 1 + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3})

     = ((x + 1)² - (\sqrt{2})²)((x² - (\sqrt{3})²)

     = (x² + 2x + 1 - 2)(x² - 3)

     = (x² + 2x - 1)(x² - 3) ← distribute

     = x^{4} - 3x² + 2x³ - 6x - x² + 3

     = x^{4} + 2x³ - 4x² - 6x + 3

6 0
3 years ago
Please help me with the question
nikdorinn [45]

Answer:

4p,q+r

4p(q+r)

4pq+4pr

3 0
2 years ago
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