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3 years ago

For which pair of points can you use this number line to find the distance?

Mathematics

1 answer:

Brilliant_brown [7]3 years ago

3 0

Answer:

(2,0) and (2,3)

Step-by-step explanation:

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A $525 speaker now sells for $393.75. Find the percent of change.

antiseptic1488 [7]

Answer: 25%

Step-by-step explanation:

595 - 393.75 = 131.25

131.25 (amount of change)/ 525 (original amount) x 100%

= 25%

8 0

3 years ago

) A store is having a 30% off sale. Michelle uses the expression 0.7p and Stuart uses the expression p-0.3p. Who is using the co

Aleonysh [2.5K]

Answer:michelle

Step-by-step explanation: it is 70% of the cost since it’s 30% of the original

3 0

3 years ago

What is the answer for -5x-4=?

slamgirl [31]

If you want to solve for x
-5x-4=?
add 4 to both sides
-5x=4+?
divide both sides by -5
x=(4+?)/-5
x=-(4+?)/5

3 0

3 years ago

Read 2 more answers

Pablo folds a straw into a triangle with side lengths of 4x2−3 inches, 4x2−2 inches, and 4x2−1 inches. Which expression can be u

Maksim231197 [3]

Answer:

A.12x2 −6; 21 inches

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago