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Shtirlitz [24]
3 years ago
11

Need answers for a b and c, please help

Mathematics
2 answers:
rewona [7]3 years ago
7 0
B is F and C is to the right 2 and down 3
Art [367]3 years ago
3 0

Answer:

it will be the x aices

Step-by-step explanation:

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-9 - (-7) = ?<br><br> Someone please explain
castortr0y [4]
-9-(-7)
-9+7=-2
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3 years ago
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Help lolz- Acellus Write 0.34 as a fraction using the steps below
Alexxandr [17]

Answer:

Step-by-step explanation:

3.1x10 = 31, and 9x10=90, so the fraction is 31/90, whcih equals 3.44444444 . . .

4 0
3 years ago
Investigate the difference between compounding annually and simple interest
Gennadij [26K]

Step-by-step explanation:

Simple interest formula

A = P (1 + rt)

Compound interest formula

A = P(1 + \frac{r}{n})^{nt}

a.

A = 5000 (1 + 0.025*1)\\A=5000(1.025)\\A=5125

Simple interest is $125

b

. A = 5000 (1 + \frac{0.025}{1})^{1*1}      \\A=5000(1.025)\\A= 5125

Compound interest is $125

c. the result for both a and b are the same

d.

A = 5000 (1 + 0.025*3) \\A=5000(1.075) \\A=5375

the simple interest is $375

e

. A = 5000 (1 + \frac{0.025}{1})^{1*3}] \\A=5000(1.025)^3 \\A=5000(1.077)\\A= 5385

the compound interest is $385

f. the result compared, compound interest is $10 more than simple interest

g.

A = 5000 (1 + 0.02*6) \\A=5000(1.12) \\A=5600

the simple interest is $600

h.

A = 5000 (1 + \frac{0.02}{1})^{1*6}] \\A=5000(1.12)^6 \\A=5000(1.9738) \\A= 9869

the compound interest is $4869

i. the result from g and h, h is over 8 times bigger than g.

j. interest compound annually is not the same as simple interest, only for the case of a and b seeing that it is for 1 year. but for 2years and above there is difference as seen in c to h

6 0
3 years ago
In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

3 0
3 years ago
Graph the equation by translating y = |x|.
kirill [66]

Answer:

graph d it's your answer plz make me brainy least

5 0
3 years ago
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