All categories

2 years ago

I need answer for this problem

Mathematics

1 answer:

Snezhnost [94]2 years ago

3 0

Answer:

(Reflect across the line y = x then reflect across the y axis then translate 2 units in left and 5 units

down) that's the answer

Step-by-step explanation:

Hope it helps

You might be interested in

Noelle and karina both leave the Internet cafe at the same time, but in opposite directions. If karina travels 9 mph faster than

Vesna [10]

Noelle goes east at N mph.
Karina travels west at N + 9
then their velocity between them is (N + N + 9)
After 8 hours they travel 8(2N + 9) = 232
(2N + 9) = 29
2N = 20
N = 10 mph for Noelle
N+9=19 mph for Karina

7 0

3 years ago

3 of 5<br> What is the nth term rule of the linear sequence below?<br> 13,7,1, -5, -11, ...<br> © T=

OverLord2011 [107]

Answer:

The nth term is 6 .

13 — 7 = 6

7— 4 = 6

1 — (–5) = 6

-5— (-11) = 6

3 0

2 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

4^338^37 / 4^158^21 A. 4^18 8^16 B. 8^48 16^58 C. 16^18 64^16 4^18 8^58

Rainbow [258]

If the problem is supposed to read (4^33*8^37)/(4^15*8^21), then...

(4^33*8^37)/(4^15*8^21) = (4^33/4^15)*(8^37/8^21)
(4^33*8^37)/(4^15*8^21) = 4^(33-15)*8^(37-21)
(4^33*8^37)/(4^15*8^21) = 4^18*8^16

Answer: Choice A) 4^18*8^16

6 0

3 years ago

Rational number between -1 and 1<br>

Nonamiya [84]

Answer:

$\dfrac{-1}{2}$