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3 years ago

Find the GCF of 15x^4 and 35x^2

Mathematics

2 answers:

borishaifa [10]3 years ago

6 0

Answer:

Step-by-step explanation:

elena55 [62]3 years ago

5 0

10, ten is the gcf that’s your answer.

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Subtract -6x2 – 2x + 5 from 2.x2 + 8.

tekilochka [14]

Look at the attached picture

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2 years ago

HELP ME PLEASE !!!!!?!!!!!?!!!!!!!!!!!

motikmotik

Answer:

First Graph (The left-most graph)

Step-by-step explanation:

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3 years ago

Solve x42=37 x 42 = 3 7 using two different strategies. Explain each strategy.

Dafna11 [192]

Given equation : $\frac{x}{42}=\frac{3}{7}$ .

Strategy 1: We can cross mutiply both sides remove fraction form.

On cross multiplication, we get

x * 7 = 3 * 42

7x = 126.

Dividing both sides by 7, we get

Strategy 2: We can find least common denominator(lcd) of both sides and multiplying both sides by that lcd to get rid denominators from both sides.

LCD of 42 and 7 is 42.

Therefore, multiplying both sides by 42, we get

$42\times \frac{x}{42}=42\times\frac{3}{7}$

x = 6 * 3

4 0

2 years ago

Please answer this for me SHOW WORK 60 points and brainliest

Sliva [168]

Answer:

OMG YES THANKS IF U EVER COMMENT ON MY QUESTION I WILL KEEP PUTTING RANDOME STUFF MWAH

Step-by-step explanation:

3 0

3 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by h and take the limit as h approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$