Write an inequality the score in a basketball game is greater than or equal to 10 points

PLS HELP!!!!ILL MARK YOU THE BRAINLISt

Shkiper50 [21]

Answer:

(x+y)/xy or (1/x + 1/y) portion of the leaves

Step-by-step explanation:

Let the total work done to rake the leaves be a for representation.

Thus,

given Maya takes x minutes to rake the leaves

thus,

work done by may in x minutes = a

dividing both side by x

work done by maya in x/x = 1 minutes = a/x

similarly

given Calra takes y minutes to rake the leaves

thus,

work done by may in y minutes = a

dividing both side by y

work done by maya in y/y = 1 minutes = a/y

__________________________________

Total work done by both in 1 minutes = a/x + a/y = a(1/x+1/y) = a(x+y)/xy

Thus, if a is the total work , then they do (x+y)/xy of a work in one minute.

Thus, (x+y)/xy portion of leaves do they rake in one minute if they work together.

8 0

2 years ago

Find the derivative

zalisa [80]

First use the chain rule; take $y=\dfrac{x+5}{x^2+3}$ . Then

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dy}\cdot\dfrac{\mathrm dy}{\mathrm dx}$

By the power rule,

$f(x)=y^2\implies\dfrac{\mathrm df}{\mathrm dy}=2y=\dfrac{2(x+5)}{x^2+3}$

By the quotient rule,

$y=\dfrac{x+5}{x^2+3}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)\frac{\mathrm d(x+5)}{\mathrm dx}-(x+5)\frac{\mathrm d(x^2+3)}{\mathrm dx}}{(x^2+3)^2}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)-(x+5)(2x)}{(x^2+3)^2}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3-10x-x^2}{(x^2+3)^2}$

So

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)}{x^2+3}\cdot\dfrac{3-10x-x^2}{(x^2+3)^2}$

$\implies\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)(3-10x-x^2)}{(x^2+3)^3}$

6 0

3 years ago

What will be the new position of the given point after rotating 180 about the origin

leva [86]

Answer: a) First quadrant

Step-by-step explanation:

From the given picture, the position of the given point is in third quadrant.

Every point in third quadrant are in the form of (-a,-b), where a and b are any positive numbers [both x and y axis are negative there]

Also after rotation of 180° the point (x,y) will map to (-x,-y)

therefore, after rotation of 180° the point (-a,-b) will map to

(-(-a),-(-b))=(a,b), where a and b are any positive numbers

Thus, the coordinate of image point =(a,b)

Since both x and y coordinate of the image are positive therefore it must lie in the first quadrant.

4 0

2 years ago

Read 2 more answers

Suppose a bag contains many marbles, 54% of which are purple. You draw 5 marbles from the bag with replacement.

dolphi86 [110]

Part A

Use the binomial probability distribution formula.

p = 0.54 = probability of getting a purple marble

n = 5 = sample size

x = 2 = number of purple we want to get

$P(x) = \frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}\\\\P(2) = \frac{5!}{2!(5-2)!}*0.54^2*(1-0.54)^{5-2}\\\\P(2) = 10*0.54^2*0.46^3\\\\P(2) = 0.283831776\\\\$

The $\frac{n!}{x!(n-x)!}$ portion is from the nCr combination formula. The exclamation marks indicate a factorial.

Alternatively, you could use Pascal's Triangle for that portion.

<h3>Answer: 0.283831776</h3>

This decimal value is exact. Round it however you need to.

============================================================

Part B

To find the expected value, aka the mean, we multiply the sample size and probability of getting a purple marble on any single selection.

n*p = 5*0.54 = 2.7

<h3>Answer: 2.7</h3>

6 0

1 year ago

How is estimating quotients different from estimating products?

tatyana61 [14]

Quotients can sometimes be irrational so it is necessary to estimate them, products on the other hand are only rational, provided they aren't multiplied by an irrational, so it is not necessary to estimate

7 0

3 years ago