All categories

3 years ago

Need help fast!!!!!!!!!!!!!!!

Mathematics

2 answers:

jenyasd209 [6]3 years ago

8 0

k(3)=13(3)-2

k(3)= 39-2

Answer is k(3)= 37

Anuta_ua [19.1K]3 years ago

6 0

ANSWER

$k(3) = 37$

EXPLANATION

The given expression is

$k(t) = 13t - 2$

We want to find k(3).

We substitute t=3 into the expression to get:

$k(3) = 13(3) - 2$

We multiply out to get,

$k(3) = 39- 2$

Simplify to get,

$k(3) = 37$

You might be interested in

The sum of 15 and three times a number is 27. Find the number.

ser-zykov [4K]

Answer:

x is a number that is multiple by 3

Step-by-step explanation:

00000009999000000p00

8 0

2 years ago

3(d-8) - 5 = 9(d+2) + 1

attashe74 [19]

Answer:

d = - 8

Step-by-step explanation:

3(d-8) - 5 = 9(d+2) + 1

3d-24 - 5 = 9d+18 + 1

3d - 29 = 9d + 19

9d - 3d = -29 - 19

6d = - 48

6d / 6 = -48/6

d = -8

7 0

3 years ago

Read 2 more answers

Please help solve for x

Ludmilka [50]

Answer:

8.49

Step-by-step explanation:

there is a little formula related to the famous formula of Pythagoras.

it says that the height of a triangle is the square root of the product of both segments of the baseline (the segments the height splits the baseline into).

so, x is actuality the height of the triangle.

x = sqrt(3×24) = sqrt(72) = 8.49

7 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

T-SHIRTS You can buy 3 T-shirts for $24. Write a proportion that gives the cost

GrogVix [38]

7:56

Figure out what 7/3 is so you can multiply 24 by that number. Is is 2 1/3 so when you multiply 2 1/3 by 24 it gets you to 56 so the proportion is 7:56

8 0

2 years ago