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fredd [130]
3 years ago
5

Help please algebra 8th grade need to finish ASAP!!

Mathematics
2 answers:
GenaCL600 [577]3 years ago
6 0
The answer is 56


hope this helps (:
igor_vitrenko [27]3 years ago
6 0
You would add 91+33 to get 124 and since all the angles in a triangle add up to 180 you would subtract 124 from 180 to get 56
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Four years ago Ravi was 8 years older than Priya. If he is now 3times as old as she is, find their present ages?
GenaCL600 [577]

Answer:

24

Step-by-step explanation:

Hope this helps and please mark me brainliest :D

4 0
3 years ago
Perform the indicated operation.(7 - 11/) + (-3 + 5/)
den301095 [7]

Answer:

<u>The correct answer is C. 4 -6</u>

Step-by-step explanation:

Let's solve the operation given, this way:

(7 - 11/) + (-3 + 5/)

7 - 11 - 3 + 5=

12 - 14 = - 2

<u>After resolving the parenthesis, adding and subtracting the result of thhe operation is - 2</u>

<u>The correct answer is C. 4 -6</u>

4 0
3 years ago
It takes Emma 18 minutes to type and spell check 14 pages of a manuscript. Find how long it takes her to type and spell check 77
NNADVOKAT [17]

Answer:

99 minutes, or an hour and 39 minutes.

Step-by-step explanation:

It takes Emma 18 minutes to do 14 pages. That means that we can set up a proportion, where Emma uses 18 minutes to do 14 pages, which is equal to x minutes per page!

\frac{18}{14} =\frac{x}{1}

14x = 18

7x = 9

x = 9/7

So, each page takes 9/7 minutes for Emma to finish.

She will do 77 pages. 77 * (9/7) = 11 * 9 = 99.

So, it will take Emma 99 minutes, or an hour and 39 minutes, to type and spell check 77 pages.

Hope this helps!

7 0
3 years ago
Read 2 more answers
Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
Element X decays radioactively with a half life of 12 minutes. If there are 200 grams of Element X, how long, to the nearest ten
Semenov [28]

Answer:

It would take 24 minutes for the element to decay to 50 grams

Step-by-step explanation:

The equation for the amount of the element present, after t minutes, is:

Q(t) = Q(0)e^{-rt}

In which Q(X) decays radioactively with a half life of 12 minutes.(0) is the initial amount and r is the rate it decreases.

Half life of 12 minutes

This means that Q(12) = 0.5Q(0)

So

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-12r}

e^{-12r} = 0.5

\ln{e^{-12r}} = \ln{0.5}

-12r = \ln{0.5}

12r = -\ln{0.5}

r = -\frac{\ln{0.5}}{12}

r = 0.05776

If there are 200 grams of Element X, how long, to the nearest tenth of a minute, would it take the element to decay to 50 grams?

This is t when Q(t) = 50. Q(0) = 200.

Q(t) = Q(0)e^{-rt}

50 = 200e^{-0.05776t}

e^{-0.05776t} = 0.25

\ln{e^{-0.05776t}} = \ln{0.25}

-0.05776t = \ln{0.25}

0.05776t = -\ln{0.25}

t = -\frac{\ln{0.25}}{0.05776}

t = 24

It would take 24 minutes for the element to decay to 50 grams

6 0
3 years ago
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