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2 years ago

A basket contains 15/16 pound of oranges .One serving of oranges is 1/8 of a pound.How many servings of oranges are there

Mathematics

1 answer:

nasty-shy [4]2 years ago

7 0

Answer:

2/15 servings

Step-by-step explanation:

IF a basket contains 15/16 pound of oranges, then;

1 servings= 15/16 pounds of oranges

To get the number of servings for 1/8 pounds

x servings = 1/8 pounds

Divide both expressions

1/x = (15/16)/(1/8)

1/x = 15/16 * 8/1

1/x = 15/2

x = 2/15

Hence there will be 2/15 servings of oranges

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3 years ago

Which shows the prime factorization of 80? Check all that apply. 2 × 4 × 10 2 × 2 × 2 × 2 × 5 24 × 5 2 × 5 × 8

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Answer:

The Prime Factorization of 80 is, 2 × 4 × 10, 2 × 2 × 2 × 2 × 5 and 2 × 4 × 10

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3 years ago

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Fresno has sunny days for 80% of the year. If there are 365 days in a normal year, how many sunny days are there each year in Fr

Alex

Answer:

l think it is 117days

Step-by-step explanation:

80%of365

80/100 of 365=116.8

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2 years ago

Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of

Doss [256]

Answer:

$\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\$

Hence, the slope , $\dfrac{dy}{dx} = \dfrac{-1}{64}$

Step-by-step explanation:

We need to find the slope, i.e. $\dfrac{dy}{dx}$ .

and all the functions are in terms of $t$ .

So this looks like a job for the 'chain rule', we can write:

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)$

Given the functions

$x = f(t)\\y = g(t)\\$

and

$x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)$

we can differentiate them both w.r.t to $t$

first we'll derivate Eq(B) to find dx/dt

$x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\$

we can also rearrange Eq(B) to find x in terms of t , $x = (37 - 4t^2)^{1/3}$ . This is done so that $\frac{dx}{dt}$ is only in terms of t.

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8$

now let's differentiate Eq(C) to find dy/dt

$2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}$

rearrange Eq(C), to find y in terms of t, that is $y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}$ . This is done so that we can replace y in $\frac{dy}{dt}$ to make only in terms of t

$\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\$

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

$\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}$

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that $\frac{dy}{dt}$ is being multiplied with $\frac{dt}{dx}$ and NOT $\frac{dx}{dt}$ , so we have to use the reciprocal!

$\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}$

our slope is equal to $\dfrac{-1}{64}$

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3 years ago