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Reptile [31]
1 year ago
10

What are the leading coefficient and degree of the polynomial? -1+23x² +8x

Mathematics
1 answer:
lyudmila [28]1 year ago
4 0
It is 12 it’s is 12 it has to be 12
You might be interested in
Solve for u.<br> U/-1 +-4=-1<br> U =
kherson [118]

Answer:

U = -3

Step-by-step explanation:

U/-1 +-4=-1

Add 4 to each side

U/-1 +-4+4=-1+4

U / -1 = 3

Multiply each side by -1

U / -1  * -1 = 3* -1

U = -3

3 0
3 years ago
Help please!!!!!!!!!!!!!
KatRina [158]
Hello!

For this you can plug in the x values in the choices and see if you get the y value

y = -4^{3}  - (-4)

-4^3 is -64

when you subtract a negative you are just adding

-64 + 4 = 60

Then you do the second one

y =  -3^{3}  - (-3)

-3^3 is -27

-27 + 3 is -24

Then the third one

y =  -2^{3}  - (-2)

-2^3 is -8

-8 + 2 is -6

Then the last one

y =  -1^{3}  - (-1)

-1^3 is -1

-1 + 1 = 0

Since we did not get -2 the answer is D

Hope this Helps!
7 0
3 years ago
Consider z = 3StartRoot 3 EndRoot + 3i. What happens to the modulus and argument when z is raised to the 4th power?
vfiekz [6]

Answer:

Its A on edge

The modulus increases by a factor of 216, and the argument increases by StartFraction pi Over 2 EndFraction.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
PLZ HELP !! Which of the points (s, t) is not one of the vertices of the shaded region of the
stira [4]

Answer:

A(0,10)

Step-by-step explanation:

Given the 4 inequalities:

s ≥ 12 - 0.5t (1)

s ≥ 10 -t (2)

s ≤ 20-t (3)

s ≥ 0

t ≥ 0

Let analyse all 4 possible answer:

  • A(0, 10)

Let substitute this point into (1) we have: 10 ≥ 12 -0.5*0 Wrong

We do not choose A

  • B (0, 20)

Let substitute this point into (1) we have: 20 ≥ 12 -0.5*0  True

Let substitute this point into (2) we have: 20 ≥ 10 - 0 True

Let substitute this point into (3) we have: 20 ≤ 20 - 0 True

We choose B as the vertex

  • C. (4, 16)

Let substitute this point into (1) we have: 16 ≥ 12 -0.5*4  True

Let substitute this point into (2) we have: 16 ≥ 10 - 4 True

Let substitute this point into (3) we have: 16 ≤ 20 - 4 True

We choose C as the vertex

  • D. (16,4)

Let substitute this point into (1) we have: 4 ≥ 12 -0.5*16 True

Let substitute this point into (2) we have: 4 ≥ 10 - 16 True

Let substitute this point into (3) we have: 4 ≤ 20 - 16 True

We choose B as the vertex

Hence, the point A(0,10)  is not one of the vertices of the shaded region of the  set of inequalities.

3 0
3 years ago
Please help whoever answers this correctly I'll mark your answer brainliest​
Oksana_A [137]

Answers:

  1. Incorrect
  2. Correct
  3. Correct

==================================================

Explanation:

When applying any kind of reflections, the parallel sides will stay parallel. Check out the diagram below for an example of this.

So PQ stays parallel to RS. Also, QR stays parallel to PS.

The statement "PQ is parallel to PS" is incorrect because the two segments intersect at point P. This letter "P" is found in "PQ" and "PS" to show the common point of intersection. Parallel lines never intersect.

4 0
2 years ago
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