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Sloan [31]
3 years ago
5

Find the midpoint of the segment with the given endpoints. (-1,-1) and (10,4)

Mathematics
1 answer:
Dimas [21]3 years ago
4 0

<em>answer \\  = (4.5 , 1.5)  \\ please \: see \: the \: attached \: picture \\ for \: full \: solution \\ hope \: it \: helps</em>

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if a, b, and c are prime numbers, do (a b) and c have a common factor that is greater than 1? (1) a, b, and c are all different
klio [65]

If a,b, and c are prime numbers, do (a*b) and c have a common factor that is greater than 1

(1) a,b, and c are all different prime numbers

(2) c≠2

1. Let's assume values of 1,3 and 5 to a, b, and c respectively

a b = 1*3 = 3

3 and 5 do not have any common factor aside 1

Let's assume values of 1,3 and 2 to a,b and c respectively

a *b = 1*3 = 3

3 and 2 does not have a common factor aside 1

2. c \neq 2

Let's assume values of 2,7 and 3 to a b and c response

a * b = 2 *7 = 14

14 and 3 does not have a common factor aside 1

learn more about of prime number here

brainly.com/question/14410795

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8 0
10 months ago
ASAPPP!!!! HELP I WILL GIVE 100POINTS AND BRAINLIST Question 9 (Essay Worth 10 points)
Luda [366]

Answer:

Part A)

The <em>x-</em>intercepts are (-1/4, 0) and (4, 0).

Part B)

The vertex is a maximum because the leading coefficient is negative.

The vertex is (1.875, 72.25).

Part C)

We can plot the zeros and the vertex, and connect them with a curve.

Step-by-step explanation:

The function given is:

f(x)=-16x^2+60x+16

Part A)

To find the <em>x-</em>intercepts of the function, set the function equal to 0 and solve for <em>x: </em>

<em />0=-16x^2+60x+16<em />

We can divide both sides by negative four:

0=4x^2-15x-4

Factor:

0=(4x+1)(x-4)

Zero Product Property:

x-4=0\text{ or } 4x+1=0

Solve for each case:

\displaystyle x=-\frac{1}{4}\text{ or }x=4

Hence, our zeros are:

\displaystyle \left(-\frac{1}{4}, 0\right)\text{ and } \left(4, 0\right)

Part B)

Note that the leading coefficient of our function is negative.

So, our function will be concave down.

Hence, our vertex will be the maximum.

The vertex is given by:

\displaystyle \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)

In this case, <em>a </em>= -16, <em>b</em> = 60, and <em>c</em> = 16.

Find the <em>x-</em>coordinate of the vertex:

\displaystyle x=-\frac{(60)}{2(-16)}=\frac{15}{8}=1.875

Substitute this back into the function to find the <em>y-</em>coordinate:

f(1.875)=-16(1.875)^2+60(1.875)+16=72.25

Hence, our vertex is:

(1.875, 72.25)

Part C)

Since we already determined the zeros and the vertex, we can plot the two zeros and the vertex and draw a curve between the three points.

The graph is shown below. Again, to do this by hand, simply plot the three points and connect them with a parabola. If necessary, we can also find the <em>y-</em>intercept.

7 0
2 years ago
Which of the following best describes the slope of the line below?
strojnjashka [21]

Answer:

the answer is D: undefined becous it's lies at Y-axis

3 0
3 years ago
Solve the equation. Show your work. 45 = 3b + 69
Cerrena [4.2K]
There should be no problem in finding the value of the unknown variable "b" in the equation given in the question.The equation is solvable for finding the value of "b" because it is the only unknown variable in the single equation that is given in the question.
45 = 3b + 69
Let us reverse both sides of the equation first. then, we get
3b + 69 = 45
3b = 45 - 69
3b = - 24
b = - (24/3)
   = - 8
So from the above deduction, we can easily conclude that the value of b in the given equation is -8.
8 0
3 years ago
Read 2 more answers
Write the standard equation for the hyperbola with the following conditions: vertices: (-2, -3) and (6, -3) foci: (-4, -3) and (
STALIN [3.7K]
Hello,

Vertices are on a line parallele at ox (y=-3)

The hyperbola is horizontal.

Equation is (x-h)²/a²- (y-k)²/b²=1

Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and  2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)

Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6

c²=a²+b²==>b²=36-4²=20

Equation is:
\boxed{ \dfrac{(x-2)^2}{16} - \dfrac{(y+3)^2}{20} =1}&#10;&#10;



8 0
2 years ago
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