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3 years ago

Solve the equation for y. X= my

Mathematics

1 answer:

Elodia [21]3 years ago

3 0

Answer:

y = x/m

Step-by-step explanation:

Please type I WILL NAME YOU BRAINLY Please measure the angles and answer it

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Lianne is saving to buy a DVD player that costs 2 1\2 times her weekly allowance. If she spends 3\4 of her allowance on other th

ahrayia [7]

Answer:

<em>She will take 10 weeks to save the money for the DVD player</em>

Step-by-step explanation:

Let w = Lianne's weekly allowance.

The DVD player costs 2 1/2 times her weekly allowance, thus

DVD cost = 5/2w

She gets w every week but spends 3/4w on other things, thus the remaining money is

Weekly savings =w - 3/4w = 1/4w

To know the number of weeks needed, we divide the DVD cost by the weekly savings:

$\displaystyle N = \frac{\frac{5}{2}}{\frac{1}{4}}$

Operating:

$\displaystyle N = \frac{5}{2}*\frac{4}{1}$

N = 10

She will take 10 weeks to save the money for the DVD player

3 0

3 years ago

Describe the process of solving the following equation: 3x+5=17<br><br> please lol

stealth61 [152]

Answer:

x = 2.3

Step-by-step explanation:

3 x + 5 = 17

- 5 = -5

------------------------

3x 7

----- = -----

3 3

-------------------------

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x = 2. 3

6 0

3 years ago

Read 2 more answers

A church window consisting of a rectangle topped by a semicircle is to have perimeter p . find the radius of the semicircle if t

Y_Kistochka [10]

Answer:

$r=\dfrac{p}{4+\pi}$

Step-by-step explanation:

Let r be the radius of the semicircle, then 2r is the width of the rectangle. Let y be the length of the rectangle. The perimeter of the window consists of two lengths, one width and length of semicircle, then

$p=2y+2r+\pi r.$

Express y:

$y=\dfrac{p-2r-\pi r}{2}.$

The area of the window is

$A=y\cdot 2r+\dfrac{1}{2}\pi r^2.$

Substitute y into the area expression:

$A=\dfrac{p-2r-\pi r}{2}\cdot 2r+\dfrac{1}{2}\pi r^2=pr-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2=pr-2r^2-\dfrac{1}{2}\pi r^2.$

Find the derivative A':

$A'=p-4r-\pi r.$

Equate A' to 0:

$p-4r-\pi r=0\Rightarrow r=\dfrac{p}{4+\pi}.$

When $r$ then $A'>0$ and the function A is increasing, when $r>\dfrac{p}{4+\pi},$ then $A'$ and the function A is decreasing. This means that at point $r=\dfrac{p}{4+\pi}$ the function A takes it maximal value (the area is maximal).