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3 years ago

20 POINTS AND BRAINIEST ANSWER

Mathematics

2 answers:

postnew [5]3 years ago

8 0

Answer:

The correct options are 1 and 3.

Step-by-step explanation:

It is given that x represents the number of small boxes and y represents the number of large boxes.

The post office has 10 small boxes and 9 large boxes in stock.

He needs at least 4 small boxes.

$4\leq x\leq 10$

He will buy no more than 15 boxes.

$6+9=15\leq 15$

$8+6=14\leq 15$

Therefore options 1 and 3 are correct.

$x+y\leq 15$

$7+10=17>15$

$9+8>15$

Therefore options 2 and 4 are incorrect.

yuradex [85]3 years ago

7 0

The correct answer is

8,6 & 6, 9

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What is 2(3a+2b-7) equivalent too?

zloy xaker [14]

Answer:

6a+4b-14

Step-by-step explanation:

For this, we have to use the distributive property.

This means we multiply each value in the parentheses by 2.

2*3a=6a

2*2b=4b

2*(-7)=-14

So our equation is:

6a+4b-14

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3 years ago

The probability of winning a new car is 1/500,000. What are the odds against winning the car?

BaLLatris [955]

The odds against winning the car is 500,000/1

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3 years ago

Write 8.34* 10^4 in standard from

Kamila [148]

Answer:

83,400

Step-by-step explanation:

Just carry the decimal point over to the right 4 places.

7 0

3 years ago

Read 2 more answers

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

2 years ago

Richard is $15 overdrawn on his checking account. He writes a check for $7. What is the balance in his checking account now?

Annette [7]

Since he’s already at -15, if he writes a check for 7 then he’s subtracting more from his account.
So, -15-7=-22

5 0

3 years ago