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mart [117]
3 years ago
11

Sandra wants to buy a new MP3 player that is on sale for $95. She has saved $73. How much morre money does she need?

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0
The answe is $22 because 95-73 is

95
73
----
22
iVinArrow [24]3 years ago
4 0
95 - 73= 22; She needs $22.

Hope this helps :)

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A line goes through the point (5,-7) and has a slope of m= -3. I need to know how to find the Y VALUE please explain the steps o
juin [17]
Well, don't you already have the t value ,-7.
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Minchanka [31]

Answer:

24

Step-by-step explanation:

41-17=24

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The coordinates of A and B are (3k, 8) and (k, -3) respectively. Given that the gradient of the line segment AB is 3, find the v
Vladimir [108]

Step-by-step explanation:

gradient = slope or several other words.

it describes how strongly a line (or tangent to a bent curve) is going up or down or ... if it is changing at all.

it is represented by the ratio

(y coordinate change / x coordinate change)

when going from one point on the line to another.

in our case, when going from A to B we have

x changes by -2k (from 3k to k).

y changes by -11 (from 8 to -3).

so, the gradient or slope is

-11/-2k = 3

11/2k = 3

11 = 3×2k = 6k

k = 11/6

A = (33/6, 8) = (11/2, 8)

B = (11/6, -3)

5 0
2 years ago
Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a
NARA [144]

x^3y^2+\sin(x\ln y)+e^{xy}=0

Differentiate both sides, treating y as a function of x. Let's take it one term at a time.

Power, product and chain rules:

\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}

=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}

=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}

=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)

=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)

=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}

=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)

=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)

=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0

Isolate the derivative, and solve for it:

\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by y to get rid of that fraction in the denominator.

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}

3 0
3 years ago
the probability of drawing a queen of hearts from a deck of cards is 1/52 if you drew one card at a time ( and put the card back
Aliun [14]

Answer:<4

Step-by-step explanation:

4 0
3 years ago
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