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2 years ago

Is it possible to complete

Mathematics

1 answer:

Mrac [35]2 years ago

8 0

Answer:

is it possible to complete

diagram be so that is not a function if so completely that round to show a relation but not a function if not justify your reasoning

yes

Step-by-step explanation:

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Which of these skills do you think you could teach someone else?

JulsSmile [24]

Writing algebraic expressions

4 0

2 years ago

Read 2 more answers

Rhonda bought five shirts from a

densk [106]

Answer:

Option C $10+5x=24+3x$

Step-by-step explanation:

Let

x ------> the cost of each shirt

we know that

The cost of 5 shirts plus $10 (amount of money that would remain) must be equal to the cost of 3 shirts plus $24 (amount of money that would remain). Considering that the cost of the shirts is the same

The linear equation that represent this scenario is

$10+5x=24+3x$

Solve for x

$5x-3x=24-10$

$2x=14$

$x=\$7$

8 0

2 years ago

Find the area of the iscoseles triangle first answer=brainliest

blsea [12.9K]

Answer:

$60cm^{2}$

Step-by-step explanation:

so first we need to find the height so by splitting the isosceles triangle into two right-angled triangles and then applying Pythagoras' Theorem to one of them.

$h^{2} =13^{2} -5^{2}$

$h^{2} =169-25$

$h^{2} =144$

$h=\sqrt{144}$

$h=12$

We now know the height of the triangle and can use this to go back and find the area of the isosceles triangle.

area of triangle = $\frac{1}{2}$ x base x height

area of triangle = $\frac{1}{2}$ x 10 x 12

area of triangle = 60

5 0

3 years ago

Read 2 more answers

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

2 years ago

126=76<br>132=32<br>208=58<br>261=?

SSSSS [86.1K]

Answer:

dude I ain't no albert einstein but I think it's 61

Step-by-step explanation:

stay groovy man

4 0

2 years ago