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3 years ago

Choose the equation of the line through (4, –5) and perpendicular to 6x + 12y = 13.

Mathematics

1 answer:

saw5 [17]3 years ago

5 0

Answer:

Kotz I don’t know the answer because I’m to lazy to do it - but I need to talk to you again... it’s me Ellis - you don’t have to rejoin the server but message me 503 333 0177.

Step-by-step explanation:

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The circumference of the inner circle is 44 ft. The distance between the inner circle and the outer circle is 2 ft. By how many

Usimov [2.4K]

You should draw a picture to visualize the problem easier.

The circumference of inner circle = 44 ft
That means the radius of the inner circle:

2*pi*r = 44
r = 44/(2pi) = 22/pi

Using the picture you hopefully drew, we see that the radius of the outer circle:
22/pi + 2

To figure out the circumference of the outer circle:
2pi*(22/pi + 2) = 44 + 4pi

So how much bigger is the the circumference of the outer circle than the inner circle?

44 + 4pi - 44 = 4pi

It is 4pi bigger or 4 * (3.14) = 12.56 feet

4 0

3 years ago

A submarine was located 200 feet below sea level. If it ascends 55 feet and then drops 87 feet, then which integer represents th

kirill [66]

Answer:

-232

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

How to solve 4n²-3n-7=0

almond37 [142]

$4n^2-3n-7=0\\ \\ a=4, \ \ b=-3 \ \ c=-7\\ \\ \Delta = b^{2}-4ac = (-3)^{2}-4*4* (-7)= 9+112=121$

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{ 3-\sqrt{ 121}}{2*4}=\frac{ 3-11}{8}= \frac{-8 }{ 8}=-1 \\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{ 3+\sqrt{ 121}}{2*4}=\frac{ 3+11}{8}= \frac{14 }{ 8}= \frac{7}{4}=1\frac{3}{4}$

4 0

3 years ago

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Nicholas has $50 to use while he is on a vacation. He decides to spend $5

Likurg_2 [28]

Answer:

days(d) = 50 ÷ 5

Explanation:

50 ÷ 5 = number of days

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3 years ago

F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

4 0

3 years ago