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mars1129 [50]
3 years ago
13

Megan bikes west to get from her apartment to school after school she bikes 8 miles north to her friends house how far is megans

apartment from her friends house, measured in a straight line

Mathematics
1 answer:
snow_tiger [21]3 years ago
4 0
The complete question: <span>Megan bikes 6 miles west to get from her apartment to school. After school, she bikes 8 miles north to her friend's house. How far is Megan's apartment from her friend's house, measured in a straight line?

So,Megan's apartment, the school, and Megan's friend's house form a right triangle with legs 6 miles and 8 miles as you can see in the attached picture; the hypotenuse of the right triangle is the straight distance is from Megan's apartment to her friend's house. To find that distance we are going to use the Pythagorean theorem:
</span>h= \sqrt{(6mi)^2+(8mi)^2}
h= \sqrt{36mi^2+64mi^2}
h= \sqrt{100mi^2}
h=10mi
<span>
We can conclude that distance between Megan's apartment and her friend's house is 10 miles.</span>

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N = x + y - n = 5<br> and part two<br> x = u + v = o<br><br> whats the answer?
Colt1911 [192]
Just substitute the value of x from second equation into first equation:
N = x + y - n = 5
N = u + v  + y - n = 5
N = 0 + y - n = 5

In short, N would be equal to y - n which is still equal to 5

Hope this helps!
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A textbook store sold a combined total of 260 chemistry and physics textbooks in a week. The number of chemistry books sold was
Lynna [10]

Let:

x = Chemistry textbooks sold

y = Physics textbooks sold

The textbook store sold a combined total of 260, so:

x+y=260

The number of chemistry books sold was three times the number of physics textbooks sold, so:

x=3y

Let:

\begin{gathered} x+y=260_{\text{ }}(1) \\ x=3y_{\text{ }}(2) \end{gathered}

Replace (2) into (1):

\begin{gathered} 3y+y=260 \\ 4y=260 \\ y=\frac{260}{4} \\ y=65 \end{gathered}

Replace the value of y into (2):

\begin{gathered} x=3(65) \\ x=195 \end{gathered}

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If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.
Lelechka [254]

Answer:

\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)

So we can say that

a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab

and we can notice that

a^2+b^2=(a+b)^2-2ab so

(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab

it comes

x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}

multiply by 3

3x^2-20x+2*16+5=3x^2-20x+37

4 0
3 years ago
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