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2 years ago

Number 5 has me going crazy, plz help

Mathematics

1 answer:

Ber [7]2 years ago

7 0

Well, the amount of degrees is 1,980, so plug that in so that the equation is 1980 = (n-2)180. Then, do order of operations. Divide 1980/180, which is 11. Then, just add 2. So, the polygon has 13 sides.

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Katerina is a botanist studying the production of pears by two types of pear trees. she noticed that Type A trees produced 20 pe

Ronch [10]

Answer:

120 pears

Step-by-step explanation:

Let the number of pears produced by Type B be x

Number of pears produced by type A=144

According to question

Type A produce 20% more pears than B

x+ x ×20%= 144

x+x× $\frac{20}{100}$ = 144

$\frac{100x+20x}{100}$ = 144

$\frac{120x}{100}$ =144

x= 144× $\frac{100}{120}$

x=120 pears

Hence, Number of pears produced by B= 120 pears

Hence, the correct answer is 120 pears

5 0

2 years ago

Write 90% as a decimal and fraction in simplest form

solong [7]

90% as a decimal is .9 and then as a fraction in simplest form is 9/10
Hope it helps

5 0

2 years ago

Factorise fully − z − 9

NeX [460]

Answer:

Yes

Step-by-step explanation:

Question

4 0

1 year ago

Solve for a: 2a + 8b = 4c

professor190 [17]

The answer is
a=2c-4b

8 0

2 years ago

Read 2 more answers

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

1 year ago