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Kisachek [45]
2 years ago
11

F(x) = 3x2+ 6x - 18 find f (10)

Mathematics
1 answer:
sesenic [268]2 years ago
8 0

Answer:

f(10) = 342

Step-by-step explanation:

f(x) = 3x² + 6x - 18

f(10 = 3(10)² + 6(10) - 18

f(10) = 3(100) + 60 - 18

f(10) = 300 + 60 - 18

f(10) = 360 - 18

f(10) = 342

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May someone please help me with this Math problem, Thank you
Semmy [17]

Answer:

c. 10

Step-by-step explanation:

Eliminate parentheses using the distributive property.

3(b+4) -2(2b+3) = -4 . . . . given

3b +12 -4b -6 = -4 . . . . . parentheses eliminated

-b +6 = -4 . . . . . . . . . . . . collect terms

-b = -10 . . . . . . . . . . . . . . subtract 6

b = 10 . . . . . . . . . . . . . . . . multiply by -1

_____

The rules of equality require that any operation you perform on one side of the equal sign must also be done on the other side. So when we say "subtract 6", we assume you know that means "subtract 6 from both sides of the equation", for example.

8 0
3 years ago
Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)
Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

x^{2} +y^{2}+4x-1=0

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

(x-h)^2+(y-k)^2=r^2

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

      (\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2}  )

    (-2,0)

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}

r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

r=\sqrt{5}

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

5 0
3 years ago
I’m really confused. Please help 。゚(゚´ω`゚)゚。
svp [43]

Answer:

4

Step-by-step explanation:

I say its 4 and it is 4

5 0
3 years ago
Help Please! Is the relationship shown by the data linear? How can you tell if it is or is not? If so, model the data with an eq
BabaBlast [244]
It is linear. The equation would be: y=2x+19
5 0
3 years ago
Read 2 more answers
Please show all work thank you math experts you’re the best !
Ne4ueva [31]

Answer:

1. -141

2. 0.98

3. 62.13

Step-by-step explanation:

1. 35 - 176 -> 176 - 35 = 141, then just make it negative

2. 0.55 x 1.79 = 0.9845 -> 0.98

3. 11.38 x 5.46 = 62.1348 -> 62.13

6 0
3 years ago
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