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kumpel [21]
2 years ago
5

Given that P = (-4, 11) and Q = (-5, 8), find the component form and magnitude of vector QP

Mathematics
1 answer:
Svetach [21]2 years ago
8 0
The given points are
P = (-4,11)
Q = (-5,8)

The x-component of vector QP is
-4 - (-5) = 1
The y-component of vector QP is
11 - 8 = 3

The vector QP is 
(1,3) or
\vec{QP} = \hat{i} + 3\hat{j}

The magnitude of the vector is
√(1² + 3²) = √(10)

Answer:
\vec{QP} = \hat{i}+3\hat{j} \,\, or \,\, (1,3)
The magnitude is √(10).

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Divide.<br><br> (6x^2+36x+35)/(x+5)<br><br> Your answer should give the quotient and the remainder.
zheka24 [161]

Answer:

6x+6+\frac{5}{x+5}

Step-by-step explanation:

When we divide polynomials by polynomials with more than one term, we use long division or factorization to simplify.

This polynomial isn't easily divided after factorization so we will use long division. And we will use the remainder theorem to write any remainder.

\frac{(6x^2+36x+35)}{(x+5)}

We start long division by finding what multiplies with x+5 to get  6x^2+36x. This is 6x.

So 6x(x+5)=6x^2+30x. We have 6x left as a remainder from 36x.

We now divide x+5 into 6x+35. What multiplies with x+5 to get 6x+35? 6.

So we have 6x+6 as our answer so far and after we multiply 6(x+5)=6x+30 we will have a final remainder of 5.

We write our answer as 6x+6+\frac{5}{x+5}.


3 0
3 years ago
If the equation x^2 + 16x + c = 0 has only one real solution, the value of c must equal
kumpel [21]

If the equation x² + 16x + c = 0 has only one real solution, the value of c is equal to 64.

<h3>Discriminant: </h3>

The quadratic formula includes the discriminant, which comes after the square root. The discriminant of a quadratic equation is important because it indicates the quantity and kind of solutions.  

The formula for the Discriminant of a Quadratic Equation ax²+bx +c = 0 is given by

<h3>               Discriminant Δ = b² - 4ac </h3>

Here we have

x² + 16x + c = 0 equation has only one real solution

Compare given equation with ax²+ bx + c = 0

=> a = 1, and b = 16  

As we know when an equation has only one real solution

then its Discriminant will be equal to zero

=> b² - 4ac = 0

By the above values,

=> 16² - 4(1)c = 0  

=> 256 - 4c = 0

=> 4c = 256

=> c = 256/4

=> c = 64

Therefore,

If the equation x² + 16x + c = 0 has only one real solution, the value of c is equal to 64.

 

Learn more about Discriminant at

brainly.com/question/15884086

#SPJ1

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1 year ago
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Answer:

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(x+1)(x+2)=0

x=-1,x=-2

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