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Rom4ik [11]
3 years ago
6

Find the values of a b and c in the table

Mathematics
2 answers:
Debora [2.8K]3 years ago
6 0

Answer: a=1, b=2, c=below x

Step-by-step explanation:

oksian1 [2.3K]3 years ago
3 0

Answer: use google calculator It’s reliable

Step-by-step explanation:

You might be interested in
Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex
siniylev [52]

Answer:

\large \boxed{\sf \ \ \ p=-11 \ \ \ }

Step-by-step explanation:

Hello,

\alpha \text{ and } \beta \text{ are the roots of the following equation}

   2x^2+6x-7=p

It means that

   2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\

And we know that

\alpha= 2\cdot \beta

So we got two equations

   2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1

For \beta =0, \ \ \alpha =0, \ \ p = -7

For \beta =-1, \ \ \alpha =-2, \ \ p= 2-6-7=-11, \ p=2*4-12-7=-11

I assume that we are after two different roots so the solution for p is p=-11

b) \alpha +2 =-2+2=0 \ and \ \beta+2=-1+2=1

So a quadratic equation with the expected roots  is

x(x-1)=x^2-x

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

3 0
3 years ago
Select ALL the correct answers.
DaniilM [7]
Be:
Number of hours: n

<span>The cost of renting a bike for the first hour is $7:
n=1→f(n)=f(1)=$7

</span>He is charged $2.50 for every additional hour of renting the bike:
f(n)=f(n-1)+2.50, for <span>n ≥ 2
</span>
f(1)=7; f(n)=f(n-1)+2.50, for <span>n ≥ 2 (sixth option)
</span>
f(n)=f(1)+2.50(n-1)
f(n)=7+2.50(n-1)
f(n)=7+2.50n-2.50
f(n)=2.50n+4.50 (fifth option)

Answers:
Fifth option: f(n)=2.50n+4.50, and
Sixth option: f(1)=7; f(n)=f(n-1)+2.50, for <span>n ≥ 2</span>
7 0
3 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
4 years ago
7. Simplify the answer to the equation. 3k + 4 + 2k +5​
nignag [31]

Answer:

5k + 9

Step-by-step explanation:

3k + 4 + 2k + 5 =

= 5k + 9

4 0
3 years ago
Read 2 more answers
2x2 + 3x ANSWER TO THIS
igomit [66]

Answer:

x(2x+3)

Step-by-step explanation:

Im guessing 2x2 is 2x^2

2x^2 + 3x = 0

x(2x+3)

3 0
4 years ago
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