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
dedylja [7]
3 years ago
11

Write a word phrase for the expression 10 + (6 – 4).

Mathematics
2 answers:
Kitty [74]3 years ago
7 0
Ten added by a subtraction of six and four
svet-max [94.6K]3 years ago
5 0

The sum of 10 and  a subtraction of 4 from 6.  

You might be interested in
How many hot dogs are eaten at major league baseball games during a season?
const2013 [10]

Answer:about 250

Step-by-step explanation:

It’s because I go to base ball games all the time and I asked before

6 0
3 years ago
Determine whether the statement is true or false. If it is true, explain why.
Over [174]
True.

Since y^4 \geq 0, y' = -1 - y^4 \ \textless \  0 and the solutions are decreasing functions.
8 0
3 years ago
Tom collects black and white model cars. he has 72 in all. he has eight more than three times as many black ones than white ones
Anarel [89]
Is this multiple choice?
4 0
3 years ago
A cylinder has a diameter of 10 inches and a height of 2.3 inches. What is the volume of this cylinder, to the nearest tenth of
Harman [31]
Cvol=hpir^2
d/2=r
10/2=5=r
h=2.3


pi=aprox 3.141592
v=2.3*3.141592*5^2
v=180.64
round to tenth
v=180.6 cubic inches
7 0
3 years ago
Observe that x and e^x are solutions to the homogeneous equation associated with:
yanalaym [24]

To take advantage of the characteristic solutions y_1(x)=x and y_2(x)=e^x, you can try the method of variation of parameters, where we look for a solution of the form

y=y_1u_1+y_2u_2

with the condition that

{u_1}'y_1+{u_2}'y_2=0

\implies{u_1}'x+{u_2}'e^x=0 (\mathbf 1)

Then

y'={y_1}'u_1+y_1{u_1}'+{y_2}'u_2+y_2{u_2}'

\implies y'={y_1}'u_1+{y_2}'u_2

y''={y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}'

Substituting into the ODE gives

(1-x)({y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}')+x({y_1}'u_1+{y_2}'u_2)-y_1u_1+y_2u_2=2(x-1)^2e^{-x}

Since

y_1=x\implies{y_1}'=1\implies{y_1}''=0

y_2=e^x\implies{y_2}'=e^x\implies{y_2}''=e^x

the above reduces to

(1-x)({u_1}'+e^x{u_2}')=2(x-1)^2e^{-x}

{u_1}'+e^x{u_2}'=2(1-x)e^{-x} (\mathbf 2)

(\mathbf 1) and (\mathbf 2) form a linear system that we can solve for {u_1}',{u_2}' using Cramer's rule:

{u_1}'=\dfrac{W_1(x)}{W(x)},{u_2}'=\dfrac{W_2(x)}{W(x)}

where W(x) is the Wronskian determinant of the fundamental system and W_i(x) is the same determinant, but with the i-th column replaced with (0,2(x-1)^2e^{-x}).

W(x)=\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=e^x(x-1)

W_1(x)=\begin{vmatrix}0&e^x\\2(x-1)^2e^{-x}&e^x\end{vmatrix}=-2(x-1)^2

W_2(x)=\begin{vmatrix}x&0\\e^x&2(x-1)^2e^{-x}\end{vmatrix}=2xe^{-x}(x-1)^2

So we have

{u_1}'=\dfrac{-2(x-1)^2}{e^x(x-1)}\implies u_1=2xe^{-x}

{u_2}'=\dfrac{2xe^{-x}(x-1)^2}{e^x(x-1)}\implies u_2=-x^2e^{-2x}

Then the particular solution is

y_p=2x^2e^{-x}-x^2e^{-x}=x^2e^{-x}

giving the general solution to the ODE,

\boxed{y(x)=C_1x+C_2e^x+x^2e^{-x}}

5 0
3 years ago
Other questions:
  • How to solve <br> 3/8=24/c+10
    15·1 answer
  • Simplify the expression. 16 • 4–4 a. –256 b. –4,096 c. 256 d. 1 divided by 16.
    11·2 answers
  • (16+20)this is my question
    9·2 answers
  • I don’t know what i’m doing!
    6·1 answer
  • 10 points and brainliest. solve by elimination
    9·1 answer
  • Read and do assignment<br> and thanks for the help and have a nice day
    10·1 answer
  • PLEASE HELP IM BEGGING YOU
    12·1 answer
  • A standard die is rolled two times, then a letter
    15·1 answer
  • Suppose that 2 were subtracted from each of the values in a data set that
    13·1 answer
  • Work out the area of the circle.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!