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3 years ago

Do #11 and #12 please answer step by step thxx

Mathematics

1 answer:

Helga [31]3 years ago

6 0

Answer:

11. x=16

12. $\sqrt{1081}$ or about 32.88

Step-by-step explanation:

11. Since there's a square you know that the side is 12 on the triangle. You can use the Pythagorean theorem (a^2+b^2=c^2) to find x. In this case, it would be 144+x^2=400. x^2=400-144=256 so x=16.

12. You can find the shortest side of the larger triangle by using the pythagorean theorem. So it would be a^2=169-25 so a=12 then you can use the same theorem and x^2=35^2-12^2=1225-144=1081 so x= $\sqrt{1081}$ which is about 32.88

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Pls show work thank you!!!

Nastasia [14]

$|-16-x|=0\\\\\implies -16-x = 0\\\\\implies x = -16$

8 0

2 years ago

reasoning: if you know that 9 x 20 = 180, how can you use this to find 9 x 24? explain your strategy.

Lostsunrise [7]

9 * 20 = 180

24 has 4 more than that, so multiply 9 to 4 and add it to 180.

9 * 4 = 36

180 + 36 = 216

So 9 * 24 = 216

4 0

3 years ago

Read 2 more answers

help with this please

Ierofanga [76]

Answer:

see explanation

Step-by-step explanation:

Under a translation < 5, - 9 >

5 is added to the original x- coordinate and 9 is subtracted from the original y- coordinate, that is

A(1, 4 ) → A'(1 + 5, 4 - 9 ) → A'(6, - 5 )

B(2, - 2 ) → B'(2 + 5, - 2 - 9 ) → B'(7, - 11 )

C(- 3, 2 ) → C'(- 3 + 5, 2 - 9 ) → C'(2, - 7 )

7 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

What is 18 divided by 6 and add 4? lk it sounds easy but yep

snow_lady [41]

18 divided by 6 is 3

when you add 3 with 4 it gives you 7

7 is your total answer

8 0

4 years ago

Read 2 more answers