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2 years ago

Ms. Hartline grades one math test every 1⁄4 hour. How many math tests does she grade in 2 hours? HELP QUICK!!

Mathematics

1 answer:

vazorg [7]2 years ago

4 0

Answer:

8 maths tests

Step-by-step explanation:

Given data

We are told that in 1/4 hour that is 0.25hour (15 mins) She grades 1 math test

our aim is to find how many maths test she grade in 3 hours (120min)

Hence

in 15 mins she does 1 grading

in 120 mins she does x

cross multiply

15x= 120

divide both sides by 15

x= 120/15

x= 8 maths test

Hence in 2 hours, she can grade 8 maths tests

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3x+y=4 x+2y=-2 what is the answer

Taya2010 [7]

Answer:

2(3x+y=4)

1(x+2y=-2)

6x+2y=8

(-) x+2y=-2

5x=10

x=2

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2+2y=-1

-2 -2

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y=-3/2

Step-by-step explanation:

multiply the first equation by 2 to make y equal to each other.
subtract to eliminate y
divide by 5 to get x
take one of the equations
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3 years ago

Finding Derivatives Implicity In Exercise,Find dy/dx implicity. xey - 10x + 3y = 0

uranmaximum [27]

Answer:

$\dfrac{dy}{dx}=\dfrac{10-e^y}{(xe^y+3)}$

Step-by-step explanation:

Given:

The implicit equation is given as:

$xe^y-10x+3y=0$

In implicit differentiation, we treat 'y' as a function of 'x' and differentiate both sides of the equation with respect to 'x' and then collect all the $\frac{dy}{dx}$ together and finally solve for $\frac{dy}{dx}$ .

So, differentiating both sides of the above equation with respect to 'x'. This gives,

$\frac{d}{dx}(xe^y-10x+3y)=\frac{d}{dx}(0)\\\\\frac{d}{dx}(xe^y)+\frac{d}{dx}(-10x)+\frac{d}{dx}(3y)=0\\\\\textrm{Using product rule, (uv)' = uv' + vu'}\\\\(x\cdot e^y\cdot \frac{dy}{dx}+e^y\cdot 1)-10\cdot1+3\frac{dy}{dx}=0\\\\\frac{dy}{dx}(xe^y)+e^y-10+\frac{dy}{dx}(3)=0\\\\\textrm{Grouping}\ \frac{dy}{dx}\textrm{ terms together}\\\\\frac{dy}{dx}(xe^y+3)+e^y-10=0\\\\\frac{dy}{dx}(xe^y+3)=10-e^y\\\\\dfrac{dy}{dx}=\dfrac{10-e^y}{(xe^y+3)}$

Therefore, the derivative $\frac{dy}{dx}$ implicitly is:

$\dfrac{dy}{dx}=\dfrac{10-e^y}{(xe^y+3)}$

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2 years ago

Which set of ordered pairs represents a function?

mel-nik [20]

The one that doesn't have an "x" repeated, that is an X-REPEAT
notice, the pair is just an x,y pair.. .so if the "x" is repeated, is NOT a function

let's see

$\bf \{(2, -2), (\boxed{1}, 5), (-2, 2), (\boxed{1}, -3), (8, -1)\}\impliedby \textit{has an "x" rep}eated\\\\
\{(3, -1), (7, 1), (-6, -1), (9, 1), (2, -1)\}\\\\
\{(6, 8), (5, 2), (\boxed{-2}, -5), (1, -3), (\boxed{-2}, 9)\}\impliedby \textit{has an "x" rep}eated\\\\
\{(\boxed{-3}, 1), (6, 3), (\boxed{-3}, 2), (\boxed{-3}, -3), (1, -1)\}\impliedby \textit{an "x" rep}eated$

6 0

3 years ago

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Georgia [21]

Answer:

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Step-by-step explanation:

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3 years ago

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2 years ago