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3 years ago

Solve John bakes 147 donuts in 7 batches. How many does each batch contain?

Mathematics

2 answers:

lys-0071 [83]3 years ago

8 0

You can solve this by dividing the baches into the total amount of donuts.

147/7=21

Each batch has 21 donuts.

snow_tiger [21]3 years ago

6 0

21 donuts
Hope this helped

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We first write the word in math as x+3/5 = 5/7. Now we cross multiply our fractions and rewrite them with the LCD of 5 and 7. Our equation now looks like x + 21/35 = 25/35. We simplify our equation by subtracting 21/35 from both sides to get x = 4/35. 4/35 is already in lowest terms so it is our answer.

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Find a solution to y'(t) = te^-t satisfying the condition y(1) = 1.

Alex_Xolod [135]

Answer:

$y=-e^{-t}(t+1)+1+\frac{2}{e}$

Step-by-step explanation:

The given differential equation is

$y'(t)=te^{-t}$

It can be written as

$\frac{dy}{dt}=te^{-t}$

$dy=te^{-t}dt$

Integrate both sides.

$\int dy=\int te^{-t}dt$

Apply ILATE rule on right side. Here, t is first function and $e^{-t}$ is the second function.

$y=t\int e^{-t}-\int (\frac{d}{dt}t\int e^{-t})$

$y=-te^{-t}-\int (1\times (-e^{-t}))$ $\int e^{-x}=-e^{-x}+C$

$y=-te^{-t}+\int e^{-t}$

$y=-te^{-t}-e^{-t}+C$ .... (1)

Initial condition is y(1) = 1. It means at t=1 the value of y is 1.

$1=-(1)e^{-t}-e^{-(1)}+C$

$1=-e^{-1}-e^{-1}+C$

$1=-2e^{-1}+C$

$1=-\frac{2}{e}+C$

Add $\frac{2}{e}$ on both sides.

$1+\frac{2}{e}=C$

Substitute the value of C in equation (1).

$y=-te^{-t}-e^{-t}+1+\frac{2}{e}$

$y=-e^{-t}(t+1)+1+\frac{2}{e}$

Therefore, the solution of given initial value problem is $y=-e^{-t}(t+1)+1+\frac{2}{e}$ .

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3 years ago

I really need help with this question.

madreJ [45]

Answer:

The answer is D

Please mark brainliest hope it helped!

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use the remainder theorem and the factor theorem to determine whether (c+5) is a factor of (c^4+7c^3+6c^2-18c+10)

Doss [256]

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