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KIM [24]
3 years ago
8

Factor the expression and find the zeros h(x)=2x2−11x+15

Mathematics
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

Step-by-step explanation:

(2x-6)(2x-5)

2(x-3)(2x-5)

^ fully factored

x-3=0       2x-5=0

x=3          2x=5

                 x=5/2

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ANEK [815]
The last one I believe because adjacent means beside eachother
4 0
2 years ago
-1=5+x 6 how do you solve it step by step to fine the answer
Paul [167]

Answer:

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5 0
3 years ago
X²-x-6<br> What is hcf of the polynomial
Eva8 [605]

Answer:

the highest common factor of the polynomial is (x + 2) (x - 3)

Step-by-step explanation:

The computation of the highest common factor of the polynomial is as follows:

According to the question, the following equation is given

'x^2 - x - 6

Now do the factorization

x^2 -3x + 2x - 6

x(x - 3) +2 (x - 3)

(x + 2) (x - 3)

hence, the highest common factor of the polynomial is (x + 2) (x - 3)

The same is to be considered

3 0
3 years ago
a man who owned 1/4 of a business sold 1/2 of his share for 3,500 at this rate the whole business was worth​
Oliga [24]

Answer:

15

Step-by-step explanation:

3 0
3 years ago
The line l is tangent to the circle with equation x^2 + y^2=10 at the point P.
babunello [35]

Given:

The equation of a circle is

x^2+y^2=10

A tangent line l to the circle touches the circle at point P(1,3).

To find:

The equation of the line l.

Solution:

Slope formula: If a line passes through two points, then the slope of the line is

m=\dfrac{y_2-y_1}{x_2-x_1}

Endpoints of the radius are O(0,0) and P(1,3). So, the slope of radius is

m_1=\dfrac{3-0}{1-0}

m_1=\dfrac{3}{1}

m=3

We know that the radius of a circle is always perpendicular to the tangent at the point of tangency.

Product of slopes of two perpendicular lines is always -1.

Let the slope of tangent line l is m. Then, the product of slopes of line l and radius is -1.

m\times m_1=-1

m\times 3=-1

m=-\dfrac{1}{3}

The slope of line l is -\dfrac{1}{3} and it passs through the point P(1,3). So, the equation of line l is

y-y_1=m(x-x_1)

y-3=-\dfrac{1}{3}(x-1)

y-3=-\dfrac{1}{3}(x)+\dfrac{1}{3}

Adding 3 on both sides, we get

y=-\dfrac{1}{3}x+\dfrac{1}{3}+3

y=-\dfrac{1}{3}x+\dfrac{1+9}{3}

y=-\dfrac{1}{3}x+\dfrac{10}{3}

Therefore, the equation of line l is y=-\dfrac{1}{3}x+\dfrac{10}{3}.

4 0
2 years ago
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