What is 171.76 divided by 555? Pls show work

Solve for x on each of these questions please!

IgorLugansk [536]

Answer:

1. x = 10

2. x = 4

Step-by-step explanation:

I use the angle ABC method:

AB² + AC² = BC²

6² + 8² = x²

x = 10

AB² + AC² = BC²

3² + x² = 5²

x = 4

HOPE THIS HELPS AND HAVE A NICE DAY <3

5 0

2 years ago

Find a fration or mixed number that is between -3 1/4 and -3 1/8

Alenkinab [10]

A fraction could be -3 1/6

8 0

3 years ago

Solve for s. 12 - 12/18s 14 = -12

Pepsi [2]

Answer:

13224512412r21414ws

Step-by-step explanation:

homework?2312425125223

7 0

3 years ago

Factor using the identity: a3 + b3 = (a + b)(a2 – ab + b2) 8z3 + 27

vivado [14]

Answer: this is our required factor i.e.

$8z^3+27=(2z+3)(4z^2-6z+9)$

Explanation:

Since we have given that

$8z^3+27$

As we know the identity , which says that

$a^3+b^3=(a+b)(a^2-ab+b^2)$

So, we can use this here ,

$8z^3+27=(2z)^3+3^3\\\\=(2z+3)((2z)^2-2z\times 3+3^2)\\\\=(2z+3)(4z^2-6z+9)$

Hence this is our required factor i.e.

$8z^3+27=(2z+3)(4z^2-6z+9)$

7 0

3 years ago

Read 2 more answers

Find the solution of the given initial value problem: 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