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satela [25.4K]
3 years ago
6

For which function defined by a polynomial are the zeros of the polynomial -4 and -6?

Mathematics
1 answer:
V125BC [204]3 years ago
4 0
If the zeros are x=-4 and -6, then the factors are:

(x+4)(x+6) which expanded is:

f(x)=x^2+10x+24
You might be interested in
Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.
dem82 [27]

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:  

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)  

Formula used :  

→Mid point of two points- (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

→Slope of two points - \frac{y_2-y_1}{x_2-x_1})

→Perpendicular of a line = \frac{-1}{slope of line})

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB =(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)

Slope of AB =\frac{-3-3}{-4-2}=1

Slope of the bisector is the negative reciprocal of the given slope.  

So, the slope of the perpendicular bisector = -1  

Equation of AB with slope -1 and the coordinates (-1,0) is,  

(y – 0) = -1(x – (-1))  

y+x=-1………………(1)  

Similarly, for AC  

Mid point of AC = (\frac{2+2}{2},\frac{3-3}{2})=(2,0)

Slope of AC = \frac{-3-3}{2-2}=\frac{-6}{0}  

Slope of the bisector is the negative reciprocal of the given slope.  

So, the slope of the perpendicular bisector = 0  

Equation of AC with slope 0 and the coordinates (2,0) is,  

(y – 0) = 0(x – 2)  

y=0 ………………(2)  

By solving equation (1) and (2),  

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1  

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB =(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)

Slope of AB =\frac{-3-3}{-4-2}=1

Slope of the bisector is the negative reciprocal of the given slope.  

So, the slope of CF = -1  

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF  

(y – 0) = -1(x – (-1))  

y+x=-1………………(3)  

Similarly, mid point of BC =(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)

Slope of AB =\frac{-3+3}{-4-2}=0

Slope of the bisector is the negative reciprocal of the given slope.  

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))  

y+3=0

y=-3………………(4)  

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)


7 0
3 years ago
Help plz no links unhelpful answers will get reported use formulas plz. I'm being gneerous
MA_775_DIABLO [31]

Answer:

what's the question?? I b.d ont get it

7 0
3 years ago
Oceanside Bike Rental Shop charges a sixteen dollar fixed fee plus six dollars an hour for renting a bike. Dan paid fifty two do
bearhunter [10]
$52 - $16= $36
$36/6 = 6 hours
8 0
3 years ago
What value of b makes the trinomial below a perfect square x^2-bx+100
Aloiza [94]
1. Perfect square trinomials, are 2nd degree polynomials, of the form          a x^{2} +bx+c so that a \neq 0, b \neq 0, c \neq 0, which can be written as perfect squares.

2. For example (x+1) ^{2} = x^{2} +2x+1
 
(3x-1)^{2}= (3x)^{2}-2(3x)+(-1) ^{2}= 9x^{2}-6x+1    


3. Thus x^{2} +2x+1, 9x^{2}-6x+1 are perfect square trinomials.

4. x^{2} -bx+100= x^{2} -bx+ 10^{2}= (x+10)^{2}   or  (x-10)^{2}

5. In the first case -b=20, so b=-20. In the second case, -b=-20, so b=20.

6. b∈{-20, 20}
6 0
3 years ago
Hmm..... Can someone help me please ? Thanks!
kolezko [41]
I believe it is A. Hope this helps.
6 0
3 years ago
Read 2 more answers
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