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4 years ago

Need help with this math problem

Mathematics

1 answer:

kondor19780726 [428]4 years ago

6 0

I might be wrong but I think it’s ASA

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5 friends went out for lunch. their total bill was $41.25 , including tax and gratuity. one of them ordered only soda and needed

Akimi4 [234]

First of all you subtract the 2.17 dollars:
41.25-2.17=39.08
Since the other 4 people got the same meal they all have to pay the same amount so you just divide 39.08 by 4:
39.08:4=9.77

4 0

4 years ago

Please Help, Math question. What is the prime factorization of 86? Use * for multiplication.

Alecsey [184]

Answer:

2 * 43

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

You might need: CalculatorThe angle O, is located in Quadrant III, and sin((.)1213What is the value of cos((,)?Express your answ

Wewaii [24]

We know that:

$\sin (\theta_1)=-\frac{12}{13}$

There is also an interesting property that relates the sine and the cosine of an angle:

$\sin ^2(\theta_1)+\cos ^2(\theta_1)=1$

We can find the cosine of theta using this equation:

$\begin{gathered} \cos ^2(\theta_1)=1-\sin ^2(\theta_1) \\ \cos (\theta_1)=\sqrt{1-\sin^2(\theta_1)} \\ \cos (\theta_1)=\sqrt[]{1-(-\frac{12}{13})^2} \\ \lvert\cos (\theta_1)\rvert=\sqrt[]{1-\frac{144}{169}}=\sqrt[]{\frac{25}{169}} \\ \lvert\cos (\theta_1)\rvert=\frac{5}{13} \end{gathered}$

Since theta is in the third quadrant then its cosine must be a negative number so:

$\cos (\theta_1)=-\frac{5}{13}$

3 0

1 year ago

Is the inequality always, sometimes, or never true? 3x - 14 < 3(x-5)

lys-0071 [83]

Answer:

3x - 14 < 3x - 15

-14 < -15

never true

Step-by-step explanation:

8 0

4 years ago

Find LCM of: x2 - 9, 3x3 + 81

VikaD [51]

Answer:

$\boxed{\pink{\sf (x + 3) \ is\ the \ LCM . }}$

Step-by-step explanation:

Given two expressions ,

$\qquad (1)\: x^2 - 9 \\\\ \qquad (2)\:3x^3 + 81$

And , we need to find the LCM , that is lowest common factor . So , let's factorise them seperately .

Factorising x² - 9 :- 

$= x^2 - 9 \\\\= x^2 - 3^2 \\\\ \red{= (x+3)(x-3)} \qquad\bf [Using \ a^2-b^2 = (a+b)(a-b) ]$

Factorising 3x³ + 81 

$= 3x^3 + 81\\\\= 3(x^3+27) \\\\= 3( x^3+27) \\\\ = 3(x^3+3^3) \\\\ \red{= 3 [ (x+3)(x^2+9-3x) ] } \qquad \bf{[ Using \ a^3+b^3=(a+b)(a^2+b^2-ab) ] }$

Hence we can see that (x+3) is common factor in both expressions.

Hence the LCM is ( x+3 ) .

8 0

3 years ago