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3 years ago

Please help me solve this

Mathematics

1 answer:

Afina-wow [57]3 years ago

8 0

Answer:

y-27=8·(x-3)

Step-by-step explanation:

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A field has a length of 12 m and a diagonal of 13 m what is the width?

kvv77 [185]

Do Pythagorean Theorem because a rectangle divided in diagonals form right triangles
A^2+b^2=C^2
Plug in the correct numbers and solve
A and B can be switched around but C is always the longest diagonal side which is called the hypotenuse not the Adjacent sides that form a right angle
So with correct solving you should get 5.

7 0

4 years ago

Jayden bought a bike for

alexandr1967 [171]

Answer:

the total cost of the bike including modifying cost is $24,150

Step-by-step explanation:

⇒ bike cost: $21,000

⇒ 15% of $21,000 is 0.15 x 21,000 = 3,150

⇒ initial + additional cost: $21,000 + $3,150 = $24,150

AND WELCOME TO BRAINLY!!

3 0

2 years ago

Compute and simplify 4 /9 + 13/18 show work

Darya [45]

Answer:

$1\frac{1}{6}$

Step-by-step explanation:

so first find the common denominator which is 162

so on 4/9 multiply 18 to denominater and numerator which makes it 72/162

next multiply 9 to numerator and denominator in 13/18 which makes it 117/162

so now add:

$\frac{72}{162} +\frac{117}{162} = \frac{189}{162}$

so now simplification.

189/162 can be reduced by 3

63/54 is our new answer. now we can reduce it by 9

7/6 is our new answer. now we must change it to a mixed number which is $1\frac{1}{6}$

hope this helped!

6 0

4 years ago

(-4, 5) and (8, -1) y intercept

ozzi

Answer:

what is it that you are looking for tho

Step-by-step explanation:

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