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3 years ago

Need help to solve this ASAP #11

Mathematics

1 answer:

Ulleksa [173]3 years ago

8 0

The second option because angle P is very small and for angle R, it’s very wide. Angle T is a perfect 90 degree right angle

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HI can someone do these 5 problems? it will really help me out I'll give out brainliest to the person who answers them. (No link

Softa [21]

Answer:

$73.60

$345

simple interest = amount deposited x time x interest rate

600 + (600 x 0.055 x 5) = $765

600 + (600 x 0.055 x 5) > $2000

$765 $2000

He would not have $2000 in 5 years

Step-by-step explanation:

Total cost of items purchased = $75 + (2 x $8.50) = 92

If there is a 20% discount, he would pay (100 - 20%) 80% of the total cost =

0.8 x $92 = $73.60

commission earned = percentage commission x amount of sales

10% x $3450

= 0.1 x 3450 = 345

Amount he would have in his account = amount deposited + simple interest

simple interest = amount deposited x time x interest rate

600 x 0.055 x 5 = $165

Amount in his account in 5 years = $165 + 600 = $765

He would have less than $2000 in his account. he would have $765

4 0

3 years ago

(2.4-(0.3-3.21)/2+0.44/(-2))/2/5

Taya2010 [7]

The answer would be 0.3635

7 0

4 years ago

Read 2 more answers

In a survey 7/8 of the people surveyed said that recycling is important. Only 1/2 of them buy products made with recycled materi

morpeh [17]

Answer:

only 3.5 buy products made with recycled material but you have to round it to 4

5 0

3 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago

[PLEASE HELP] Consider this function, f(x) = 2X - 6.

ira [324]

Answer:

Find answer below

Step-by-step explanation:

f(x)=2x-6

Domain of 2x-6: {solution:-∞<x<∞, interval notation: -∞, ∞}

Range of 2x-6: {solution:-∞<f(x)<∞, interval notation: -∞, ∞}

Parity of 2x-6: Neither even nor odd

Axis interception points of 2x-6: x intercepts : (3, 0) y intercepts (0, -6)

inverse of 2x-6: x/2+6/2

slope of 2x-6: m=2

Plotting : y=2x-6

7 0

4 years ago