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3 years ago

7

Select all equations that could be used to find the measure of angle C, using the fact that the sum of all the angles in a trian

gle is equal to 180

1 answer:

Semmy [17]3 years ago

4 0

C = 180 - A - B

And

C = 180 - (A+B)

Hope I can help you :)
Brainliest answer :)) ?

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Explain to another student how to solve the equation.

AveGali [126]

Answer:

Solve for X ( The question I did was an example )

Step-by-step explanation:

To solve and equation like 2x+9=30 you will have to isolate the number with a variable. You need to subtract 9 on both sides for that to happen.

2x=21

Now the next step is to divide. You have to divide to get the answer.

x=21/2

x= 10.5

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2 years ago

Please help me... I really need it.

MAXImum [283]

Answer:

If you are writing a seesya let me help you

Step-by-step explanation:

Make sure to hilight key info

Also use grammarly to correct past tense and sentences

Make sure to make sense

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2 years ago

How are proptions and percents the same

Lelechka [254]

Answer:

1 would be 10%, but a proportion is a ratio, as in this case 10/100 = 1/10= . 1, but 10% is not a ratio in of itself, but is a proportion of 100. % means OF one hundred. A proportion is a number TO another number.

Step-by-step explanation:

4 0

2 years ago

Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex

siniylev [52]

Answer:

$\large \boxed{\sf \ \ \ p=-11 \ \ \ }$

Step-by-step explanation:

Hello,

$\alpha \text{ and } \beta \text{ are the roots of the following equation}$

$2x^2+6x-7=p$

It means that

$2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\$

And we know that

$\alpha= 2\cdot \beta$

So we got two equations

$2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1$

For $\beta =0, \ \ \alpha =0, \ \ p = -7$

For $\beta =-1, \ \ \alpha =-2, \ \ p= 2-6-7=-11, \ p=2*4-12-7=-11$

I assume that we are after two different roots so the solution for p is p=-11

b) $\alpha +2 =-2+2=0 \ and \ \beta+2=-1+2=1$

So a quadratic equation with the expected roots is

$x(x-1)=x^2-x$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

3 0

2 years ago

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