1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Olenka [21]
3 years ago
12

Solve each equation for the ind2. 2(x-7)=10​

Mathematics
2 answers:
SVETLANKA909090 [29]3 years ago
3 0

Answer:

x= 12

Step-by-step explanation:

2(x-7) = 10

Distribute the 2. 2x-14 = 10

Add the 14 onto both sides to cancel it out on the original side and following the rules- the same must be done to the other side. 2x-14 (+14) = 10+14 -----> 2x = 24

Divide both sides by 2 to get the X alone. 2x/2 = 24/2 ---------> x= 12

ch4aika [34]3 years ago
3 0

Answer:

x=12

Step-by-step explanation:

You might be interested in
Answer fast please and thank you
steposvetlana [31]

Answer:

Yes

Step-by-step explanation:

One input to one output is a linear one-to-one function.

7 0
2 years ago
Read 2 more answers
Which of the following best describes the graphs below?
Sati [7]

Answer:

D

Step-by-step explanation:

Look at all possibilities then use  all of them then find out which one matches the best.

8 0
3 years ago
5. Susan receives a Federal Direct Subsidized Loan for $1,000 as a freshman in
PSYCHO15rus [73]

Answer:

c

Step-by-step explanation:

subsidized means that they pay for it

4 0
2 years ago
Find the laplace transform by intergration<br> f(t)=tcosh(3t)
Shkiper50 [21]
\mathcal L_s\{t\cosh3t\}=\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt

Integrate by parts, setting

u_1=t\implies\mathrm du_1=\mathrm dt
\mathrm dv_1=\cosh3t e^{-st}\,\mathrm dt\implies v_1=\displaystyle\int\cosh3t e^{-st}\,\mathrm dt

To evaluate v_1, integrate by parts again, this time setting

u_2=\cosh3t\implies\mathrm du_2=3\sinh3t\,\mathrm dt
\mathrm dv_2=\displaystyle\int e^{-st}\,\mathrm dt\implies v_2=-\frac1se^{-st}

\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\int \sinh3te^{-st}

Integrate by parts yet again, with

u_3=\sinh3t\implies\mathrm du_3=3\cosh3t\,\mathrm dt
\mathrm dv_3=e^{-st}\,\mathrm dt\implies v_3=-\dfrac1se^{-st}

\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\left(-\frac1s\sinh3te^{-st}+\frac3s\int\cosh3te^{-st}\,\mathrm dt\right)
\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}+\frac9{s^2}\int\cosh3te^{-st}\,\mathrm dt
\displaystyle\frac{s^2-9}{s^2}\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}
\implies\displaystyle\underbrace{\int\cosh3te^{-st}\,\mathrm dt}_{v_1}=-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}

So we have

\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt=u_1v_1\big|_{t=0}^{t\to\infty}-\int_0^\infty v_1\,\mathrm du_1
=\displaystyle-\frac{t(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\bigg|_{t=0}^{t\to\infty}-\int_0^\infty \left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\,\mathrm dt
=\displaystyle\frac1{s^2-9}\int_0^\infty(s\cosh3t+3\sinh3t)e^{-st}\,\mathrm dt

We already have the antiderivative for the first term:

\displaystyle\frac s{s^2-9}\int_0^\infty \cosh3te^{-st}\,\mathrm dt=\frac s{s^2-9}\left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}
=\dfrac{s^2}{(s^2-9)^2}

And we can easily find the remaining term's antiderivative by integrating by parts (for the last time!), or by simply exchanging \cosh with \sinh in the derivation of v_1, so that we have

\displaystyle\frac3{s^2-9}\int_0^\infty\sinh3te^{-st}\,\mathrm dt=\frac3{s^2-9}\left(-\frac{(s\sinh3t+3\cosh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}
=\dfrac9{(s^2-9)^2}

(The exchanging is permissible because (\sinh x)'=\cosh x and (\cosh x)'=\sinh x; there are no alternating signs to account for.)

And so we conclude that

\mathcal L_s\{t\cosh3t\}=\dfrac{s^2+9}{(s^2-9)^2}
8 0
3 years ago
What is the argument of -4+4i
zhuklara [117]

the agrument of any complex number is defined as its angle.

if a complex number be Z=a+ib then the agrument of Z is given by,

Agr|Z|=A

where, tanA=b/a i.e. A=tan^-1(b/a)

by the question,

a=-4 and b=4

so A= tan^-1(4/-4)= tan^-1(-1)=135°

therefore the agrument of -4+4i is 135°

3 0
3 years ago
Other questions:
  • If anyone could help that would be great / right answer gets brainilest
    10·2 answers
  • Ginger's cat gave birth to a kitten that weighed 3 3/8 ounces when it was born. On the day Ginger sold the kitten to its new own
    11·2 answers
  • The figure shows a circle circumscribed around a triangle. what is constructed first when creating the circle?
    8·2 answers
  • How to find the area of each face of a prism
    11·1 answer
  • The sum of two integers is 9 and the sum of their squares is 53. Find the integers.
    9·1 answer
  • Please help me I need to get this right asap
    14·1 answer
  • Help please. it would really help if someone explained
    12·1 answer
  • What percent of 120 is 90
    14·1 answer
  • Shani borrowed $5000 at an annual interest rate of 4% compounded yearly. William borrowed the same amount of money at an annual
    15·1 answer
  • Uh Expanded form pls? 4.02 x 10^5
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!