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Nesterboy [21]
3 years ago
7

5(1+4K)+7k=4(7k+2) Help

Mathematics
1 answer:
stepladder [879]3 years ago
5 0

Answer:

k = -3

Step-by-step explanation:

1) Simplify

5+20k+7k=28k+8

2) Combine the k's

5+27k=28k+8

3) Subtract 27k on both sides

5=k+8

4) Subtract 8 on both sides

-3=k or k= -3

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The graph shows the career interests of the students at Linda's college. Suppose there are 900 students at the school. How many
nevsk [136]

Answer:

117 students

Step-by-step explanation:

13/100 = X/900

(13 * 900)/100 = 117

5 0
2 years ago
Review the expression below -2(a+b-5)+3(-5a+2b)+b(6a+b-8)
Oduvanchick [21]

Answer:

6ab + b^2 - 17a - 4b + 10

Step-by-step explanation:

-2(a+b-5)+3(-5a+2b)+b(6a+b-8)

Now we break the parenthesis. To break that, we multiply each of the value inside the parenthesis by the adjacent number. That is, for the first part of the expression, we multiply by -2, then by 3, and then by b.

Algebraic Operations need to be considered:

[ (-) x (-) = (+); (-) x (+) = (-)]

= [-(2*a) + (-2*b) - (-2*5)] + [3*(-5a) + (3*2b)] + [(b*6a) + (b*b) - (b*8)]

= -2a - 2b +10 -15a + 6b + 6ab + b^2 - 8b

Now, we will make the adjustment by the similarity value.

= - 2a - 15a - 2b + 6b - 8b + 6ab + b^2 + 10

= - 17a - 4b + 6ab + b^2 + 10

= 6ab + b^2 - 17a - 4b + 10

Therefore, the answer of the expression is = 6ab + b^2 - 17a - 4b + 10

6 0
3 years ago
If -1 is one of the zeroes of py 3x^3-5x^2-11x-3 find other zeroes
katrin2010 [14]

Given:

The function is:

y=3x^3-5x^2-11x-3

It is given that -1 is a zero of the given function.

To find:

The other zeroes of the given function.

Solution:

If c is a zero of a polynomial P(x), then (x-c) is a factor of the polynomial.

It is given that -1 is a zero of the given function. So, (x+1) is a factor of the given function.

We have,

y=3x^3-5x^2-11x-3

Split the middle terms in such a way so that we get (x+1) as a factor.

y=3x^3+3x^2-8x^2-8x-3x-3

y=3x^2(x+1)-8x(x+1)-3(x+1)

y=(x+1)(3x^2-8x-3)

Again splitting the middle term, we get

y=(x+1)(3x^2-9x+x-3)

y=(x+1)(3x(x-3)+1(x-3))

y=(x+1)(3x+1)(x-3)

For zeroes, y=0.

(x+1)(3x+1)(x-3)=0

(x+1)=0 and 3x+1=0 and x-3=0

x=-1 and x=-\dfrac{1}{3} and x=3

Therefore, the other two zeroes of the given function are -\dfrac{1}{3} and 3.

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