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3 years ago

Kayla works for 7 hours at $10.25 per hour. How much does he earn?

Mathematics

1 answer:

Sphinxa [80]3 years ago

8 0

Answer:

$71.75

Step-by-step explanation:

10.25 · 7 = 71.75

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Someone help me with this please!!! Urgent!!!

vazorg [7]

Answer:

Output would be f(x)

f(x) = x/2

Step-by-step explanation:

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2 years ago

PLEASE HELP FOR BRAINLIEST ANSWER NEED TO KNOW IF IM CORRECT

Juliette [100K]

You're wrong its letter a

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2 years ago

Elena starts her new job working at an office supply store. She works for 6 hours and makes a total of $54.

docker41 [41]

Answer:

I believe the answer would be A.

Step-by-step explanation:

A says 6 x h is 54.

^ ^ ^

important!

6 is the number of hours.

h is how much money made per hour.

54 is the total after working 6 hours.

4 0

2 years ago

A rock climber descended 50 feet to a new elevation of 32 feet. What was her initial elevation, e?

alina1380 [7]

Answer:

82 feet

Step-by-step explanation:

50 feet + 32 feet = 82 feet

6 0

3 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

7 0

2 years ago