All categories

4 years ago

Someone please please help a girl out

Mathematics

1 answer:

Katen [24]4 years ago

7 0

180-65=115
115-42=73
i think the answer is 73°

You might be interested in

A students pay of $21 an hour six dollars more than twice the amount the student as per hour and internship enter in solving equ

Tanya [424]

That makes no sense I think you’re missing some words but the answer might be $7.5 per hour bc 21-6 is 15 and that divided by 2 s 7.5

5 0

3 years ago

Read 2 more answers

Adante begins to evaluate the expression 3 1/3 x 5 1/4 using the steps below

il63 [147K]

Answer:

$\frac{35}{2}$

Step-by-step explanation:

To solve this problem we need to write the mixed fraction as a fractional number, as follows:

$3 1/3 = 3 + \frac{1}{3} = \frac{9+1}{3} = \frac{10}{3}$

$5 1/4 = 5 + \frac{1}{4} = \frac{20+1}{4} = \frac{21}{4}$

Then, evaluating the expression:

$\frac{10}{3}$ × $\frac{21}{4}$ = $\frac{210}{12}$ = $\frac{35}{2}$

4 0

3 years ago

Please help! ~Find the volume of the triangular prism~

12345 [234]

Answer:

23.04 m^2

Step-by-step explanation:

The volume of a prism is base * height. In the case of a triangular prism, you need to find the area of the base first (aka the triangle).

The area of a triangle is base * height / 2. Since we have the values for those we can substitute the values into the formula and solve.

2.4 * 3.6 / 2 = 3.84.

Now multiply that by height of the prism, which is 6, and we get 23.04 m^2.

Hope this helps!

7 0

3 years ago

Help me pls pls pls pls pls

Oxana [17]

Answer:

B.x>4is a required interval.

5 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