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Sever21 [200]
3 years ago
5

Find value of a if x-a is a factor of x³-ax²+2x+a-1​

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0
Your answer to this question will be a=1/3
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Solve and graph
oksano4ka [1.4K]

<span><span><span>2r - 9 > -6
</span><span>2r - 9 = -6
</span>2r = 3</span><span>
r = 3/2 = 1.5</span></span><span><span>
r > 1.5</span></span>

<span><span /></span><span><span>9x-5 < -41
</span><span>9x-5 = -41
9x = -36
x = -36/9 = -4
x < -4</span></span>

<span><span>3x + 13 > 7
3x + 13 = 7
3x = -6
x = -6/3 = -2
x > -2</span></span>

<span><span>4x + 3 > -17
4x + 3 = -17
4x = -20
x = -20/4 = -5
x > -5</span></span>

<span><span>7x - 4 < 10
7x - 4 = 10
7x = 14
x = 14/7 = 2
x < 2</span></span><span>
</span>

4 0
3 years ago
What is the answer to this
zalisa [80]

Answer:

see below

Step-by-step explanation:

The vertical asymptote is the up and down value it does not reach

y =0

The horizontal asymptote is the side  value it does not reach

x =0

4 0
3 years ago
Create a model to show 2 × 5.
Sliva [168]

Answer:

The model is shown in the attached picture. The solution is 10.

Step-by-step explanation:

There are two way of showing a model for 2×5

  1. The tape diagram has 2 equal-sized parts of 5
  2. The tape diagram has 5 equal-sized parts of 2

Both models give the same solution:

  1. 5+5=10
  2. 2+2+2+2+2=10

In the attached picture we can show two types of diagram.

5 0
3 years ago
20 points
gavmur [86]

Answer:

Michael buys 800 hockey cards from jackson

6 0
2 years ago
Read 2 more answers
Which of the following is equal to the expression above
EleoNora [17]

Answer:

15 \sqrt[3]{2}

Step-by-step explanation:

{(27 \times 250)}^{ \frac{1}{3} }  =  {(27 \times 125 \times 2)}^{ \frac{1}{3} }  \\  =  {27}^{ \frac{1}{3} }  \times  {125}^{ \frac{1}{3} }  \times  {2}^{ \frac{1}{3} }  \\  =  \sqrt[ 3]{27}  \times  \sqrt[3]{125}  \times  \sqrt[3]{2}  \\  =  \sqrt[3]{ {3}^{3} }  \times  \sqrt[3]{ {5}^{3} }  \times  \sqrt[3]{2}  \\  = 3 \times 5 \times  \sqrt[3]{2}  \\  = 15 \sqrt[3]{2}

4 0
3 years ago
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