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irina1246 [14]
3 years ago
15

1)If (x-2) is a factor of x2+ 3ax -2a, then find the value of a

Mathematics
1 answer:
liraira [26]3 years ago
5 0

Answer:

x - 2 => x = 2

x² + 3ax - 2a

=> 2² + 3 × a × 2 - 2 × a

=> 4 + 6a - 2a

=> 4 + 4a

=> 4 ( 1 + a )

★ 1 + a = 0

a = -1

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Step-by-step explanation: cause it it you’re welcome

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3 years ago
The pool at the park is 18 feet wide how many yardsticks would it take to measure the width of the pool
sattari [20]
6 because 18 divided by 3 equals 6

4 0
3 years ago
Solve 3x+2y=7 and x-4y=-21 by using any method
kupik [55]

Answer:

(- 1, 5 )

Step-by-step explanation:

Using the method of substitution.

Given the 2 equations

3x + 2y = 7 → (1)

x - 4y = - 21 → (2)

Rearrange (2) expressing x in terms of y by adding 4y to both sides.

x = 4y - 21 → (3)

Substitute x = 4y - 21 into (1)

3(4y - 21) + 2y = 7 ← distribute and simplify left side

12y - 63 + 2y = 7

14y - 63 = 7 ( add 63 to both sides )

14y = 70 ( divide both sides by 14 )

y = 5

Substitute y = 5 into (3)

x = (4 × 5) - 21 = 20 - 21 = - 1

Solution is (- 1, 5 )

8 0
4 years ago
when nancy started middle school, she was 60 inches tall. when she started high school, she was 66 inches tall. what was the per
Ivahew [28]

Answer:

The percent change in Nancy height is 10%

Step-by-step explanation:

Given : when Nancy started middle school, she was 60 inches tall. when she started high school, she was 66 inches tall.

We have to find the percent change in her height.

Percentage change is given by the ratio of the difference in old value and new value to the old value times 100

Mathematically written as

Percentage\ change =\frac{new\ value-old\ value}{Old\ value}\times 100

For the given data,

Old value = 60 inches

New value = 66 inches

Thus, the Percentage change is given as,

Percentage\ change =\frac{66-60}{60}\times 100

Simlify, we have,

Percentage\ change =\frac{6}{60}\times 100

Thus, The percent change in her height is 10%

7 0
3 years ago
Read 2 more answers
These roots of the polynomial equation x^4-4x^3-2x^2+12x+9=0 are 3,-1,-1. Explain why the fourth root must be a real number. Fin
Alex787 [66]

Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.

Because we know 3 and -1 (multiplicity 2) are both roots, the last root r is such that we can write

x^4-4x^3-2x^2+12x+9=(x-3)(x+1)^2(x-r)

There are a few ways we can go about finding r, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be (-3)(1)(1)(-r)=3r.

Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

3r=9\implies r=3

so the missing root is 3.

Other things we could have tried that spring to mind:

- three rounds of division, dividing the quartic polynomial by (x-3), then by (x+1) twice, and noting that the remainder upon each division should be 0

- rational root theorem

4 0
3 years ago
Read 2 more answers
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