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3 years ago

The point (3, 18) is on the parabola y =

Mathematics

1 answer:

liberstina [14]3 years ago

8 0

Answer:

$y = {ax}^{2} \\ 18 = (a \times {3}^{2} ) \\ 18 = 9a \\ a = 2$

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Can someone help me with this? I'm confused

lakkis [162]

Answer:

The Answer is B

Step-by-step explanation

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3 years ago

What is the area of this composite figure?

Mkey [24]

We have two congruent trapezoids.

The formula of an area of a trapezoid:

$A=\dfrac{base_1+base_2}{2}\cdot\ height$

We have:

$base_1=210\ mm\\\\base_2=150\ mm\\\\height=55\ mm$

Substitute:

$A=\dfrac{210+150}{2}\cdot55=\dfrac{360}{2}\cdot55=180\cdot55=9,900\ mm^2$

Answer:

$2A=2\cdot9,900\ mm^2=19,800\ mm^2$

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3 years ago

Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3

Whitepunk [10]

If $f(x)=4x-3$ :

$\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4$

If $f(x)=4x^{-3}$ :

$\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}$

$\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}$

$\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}$

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4 years ago

PLEASE ANSWER if u answer this without answering questions i am reporting you

Ray Of Light [21]

Answer:

Q1. m=8

Q2. Divide by -5

Q3. It's being divided by 5

Q4. It is being multiplied by -5

3 0

3 years ago

Read 2 more answers

Which number can each term of the equation be multiplied by to eliminate the fractions before solving?

Vikentia [17]

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6 0

4 years ago