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meriva
2 years ago
9

You have an initial amount of $55 in a savings account and you deposit an equal amount into your account each week. After 6 week

s the account holds $133. Write an equation that represents the amount yy (in dollars) of money in the account after x weeks
Mathematics
2 answers:
charle [14.2K]2 years ago
5 0

The answer is y=55+13x

I just had this in my homework and I got it wrong but this is the answer they told me was correct :)

yKpoI14uk [10]2 years ago
3 0
Y=55x
 The 55 will be a constant factor and the 133 won't.
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A person is standing exactly 36 ft from a telephone pole. There is a 30° angle of elevation from the ground to the top of the po
nika2105 [10]

Answer:

20.78feet

Step-by-step explanation:

The question made us to understand that the man is standing and also there is angle of elevation, then we need to draw a right triangle having a base equal to 36 feet with an angle from the base to the top of the pole which is 30 degrees.

tan= opposite side / adjacent side

Let height of the pole =h

Tan(30)= h/36

But tan 30degree= 1/√3

h= 36 × 1/√3

h= 20.78feet

Therefore, the height of the pole= 20.78feet

CHECK THE ATTACHMENT FOR DETAILED FIGURE

7 0
2 years ago
Read 2 more answers
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
2 years ago
The function y = (x + 4)4 is a transformation of the graph of the parent function y = x4. How is the zero of the parent function
Musya8 [376]
The answer is C but not sure about it please double check it


Thank you

I hope it’s help you
7 0
2 years ago
Read 2 more answers
after 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. what is the half-life of the sample?
coldgirl [10]
<span>(1/2)n = 0.015625 
n log 0.5 = log 0.015625 
n = log 0.5 / log 0.015625 
n = 6</span><span>24 days / 6 half-lives = 4.00 days (the length of the half-life)

</span>
6 0
2 years ago
What is the median of the following 7 scores?
dybincka [34]
First, you want to put all the numbers in order from least to greatest.

21, 33, 33, 42, 67, 79, 89

Now, you can use this song to help remember how to do these problems.

Cross off the sides till you get to the center. 1 is good, 2 is bad. (If you get two numbers in the middle) add then divide by two.

Cross off 21, then 89, then 33, then 79, then 33, then 67. Now, you're left with 42 in the middle. That is your median. Hope this helps! :)
8 0
2 years ago
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