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3 years ago

Whats 109 divided by 27 and then simplify it

Mathematics

1 answer:

ioda3 years ago

8 0

109/27 = 4.037037037

Uncertain on what exactly they want on simplified, here are a couple

Rounded (Hundredth) = 4.04

Rounded (Whole & tenth) = 4

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A flying squirrel lives in a nest that is 9 meters high in a tree. To reach a fallen acorn that is 4 meters from the base of the

Luba_88 [7]

Answer:

Flying squirrel have to glide is for 9.8 meters approximately.

Step-by-step explanation:

Here we can use Pythagoras theorem.

Height is 9 , base is 4.

We need to find the hypotenuse measure.

So, let's use Pythagoras theorem $a^{2} +b^{2}= c^{2}$

Plug in a as 9 and b 4.

$9^{2} +4^{2}=c^{2}$

Simplify the left side of the equation

$81+16=c^{2}$

$97=c^{2}$

Take square root on both sides.

c=9.8 meters.

6 0

3 years ago

Find the 12th term of the arithmetic sequence 2x + 8,6x + 5, 10x + 2, ...

lawyer [7]

Answer: The 12th term is 46x -25

Step-by-step explanation:

It is increasing by 4x and decreasing by 3 through each term, this represents the d. a1 has already been identified as 2x +8, so you just have to write down the formula and the solve form there by plugging in the 12 in (n-1) to find the answer.

8 0

2 years ago

What is the volume to number 5?

almond37 [142]

It is 4200 cm hope i helped

8 0

3 years ago

What is the complete factorization of 36y2 − 1?

stiv31 [10]

Answer:

(6y+1)(6y-1)

Step-by-step explanation:

Difference of square (You may want to search it up if you want to learn more)

4 0

3 years ago

Find the derivative of f(x)= (e^ax)*(cos(bx)) using chain rule

Vikentia [17]

$f(x) = e^{ax}\cos(bx)$

then by the product rule,

$f'(x) = \left(e^{ax}\right)' \cos(bx) + e^{ax}\left(\cos(bx)\right)'$

and by the chain rule,

$f'(x) = e^{ax}(ax)'\cos(bx) - e^{ax}\sin(bx)(bx)'$

which leaves us with

$f'(x) = \boxed{ae^{ax}\cos(bx) - be^{ax}\sin(bx)}$

Alternatively, if you exclusively want to use the chain rule, you can carry out logarithmic differentiation:

$\ln(f(x)) = \ln(e^{ax}\cos(bx)} = \ln(e^{ax})+\ln(\cos(bx)) = ax + \ln(\cos(bx))$

By the chain rule, differentiating both sides with respect to <em>x</em> gives

$\dfrac{f'(x)}{f(x)} = a + \dfrac{(\cos(bx))'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a - \dfrac{\sin(bx)(bx)'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a-b\tan(bx)$

Solve for <em>f'(x)</em> yields

$f'(x) = e^{ax}\cos(bx) \left(a-b\tan(bx)\right) \\\\ f'(x) = e^{ax}\left(a\cos(bx)-b\sin(bx))$

just as before.

4 0

3 years ago