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blsea [12.9K]
3 years ago
15

F(x)=-4x^2+9 Find f(−3)

Mathematics
1 answer:
pochemuha3 years ago
8 0

Answer:

f (-3) = 45

Step-by-step explanation:

f (x) =

4x {}^{2}  + 9

f (-3) =

(4 \times  - 3 \times  - 3) + 9

=》

36 + 9

=》

45

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finlep [7]

21 *78=

20 * 80=

1600 as an estimate


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3 years ago
Does anyone know the fraction pairs for these? Please help ASAP!
pishuonlain [190]

Answer:

dividing denominators

Step-by-step explanation:

you say for example 12/5, means 12 divides 5 is going to give something 2.4, so as the answer is gonna be 2 hole number 4/5.

3 0
3 years ago
Whyyyyy do I suck so baddddddd
oksian1 [2.3K]

Okay, so first, the quadrants on a coordinate plane go counter clockwise.

                          2                 |               1

                       __________|__________

                           3                |              4

(1/2, -1.8) Would be a little to the right of the y axis, and a little below the x axis.

So, your answer would be D) Quadrant IV

8 0
4 years ago
Megan and Suzanne each have a plant. They track the growth of their plants for four weeks. Whose plant grew at a faster rate, an
SOVA2 [1]
Megan`s plant:
Rate:  ( 7 - 4.5 ) / ( 2 - 1 ) = 2.5 in per week
Suzanne`s plant:
Rate:   ( 7 - 5 ) /  ( 2 - 1 ) = 2 in per week
Answer:
D ) Megan`s at 2.5 inches per week 

6 0
3 years ago
Read 2 more answers
Explain how to get that answer!!
ra1l [238]
We need to simplify \frac{ \sqrt{14x^3} }{ \sqrt{18x} }

First lets factor \sqrt{14x^3}

\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3}
\sqrt{14} =  \sqrt{2} \sqrt{7} by applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a} \sqrt[n]{b}
\sqrt{x^3} = x^{3/2} By applying the radical rule \sqrt[n]{x^m} = x^{m/n}

So
\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3} = \sqrt{2} \sqrt{7}x^{3/2}

Now let's factor \sqrt{18x}
By applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a}  \sqrt[n]{b},
\sqrt{18x} =  \sqrt{18} \sqrt{x}
\sqrt{18} =  \sqrt{2} * 3

So \sqrt{18x} = \sqrt{2}*3 \sqrt{x}

So  \frac{ \sqrt{14x^3} }{ \sqrt{18x} } = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}  }

We know that \sqrt[n]{x} = x^{1/n} so \sqrt{x} = x^{1/2}

We now have \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}

We know that \frac{x^a}{x^b} = x^{a-b}
So \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x

We now got \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}


We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} =   \frac{ \sqrt{7}x }{3}


All in All, we get \frac{ \sqrt{14x^3} }{ \sqrt{18x} } =  \frac{ \sqrt{7}x }{3}

Hope this helps! :D


6 0
3 years ago
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