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Mrac [35]
3 years ago
8

WILL GIVE THE BRAINLIEST AND 99 PTS Q-What is 8% of 1000

Mathematics
2 answers:
tatuchka [14]3 years ago
4 0

Answer: 80

Step-by-step explanation:

1000 * 0.08 = 0.80 = 80

vesna_86 [32]3 years ago
4 0

Answer:

your answer would be 80

Step-by-step explanation:

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It is 76°f at the 6000-foot level of a mountain, and 49°f at the 12,000-foot level of the mountain. write a linear equation to f
valina [46]

Answer:

t(x) = 103 -0.0045x

Step-by-step explanation:

We are given the following in the question:

Let the equation:

t(x) = a + bx

be the linear equation that represent temperature at an elevation x.

Temperature at 6000-foot level = 76 f

76 = a + 6000b

Temperature at 12000-foot level = 49 f

49 = a + 12000b

Solving the two equation, we get,

76-49 = -6000b\\27=-6000b\\\\b = \dfrac{-27}{6000}\\\\b = -0.0045\\76 = a + (-0.0045)6000\\76 = a -27\\a = 76 + 27\\a  =103

Thus, we can write the linear equation as:

t(x) = 103 -0.0045x

5 0
3 years ago
Just help please, im the NOT smartest
olasank [31]

Answer: 3/7

The slope is the same.

4 0
3 years ago
Read 2 more answers
Write 0.000000364 in standard form
Thepotemich [5.8K]

Answer:

yes

Step-by-step explanation:

this number is a decimal

5 0
3 years ago
Tom Thomas took tennis lessons. He traveled 12 miles one way at an average cost of $0.42 per mile. His racket cost $94, balls co
Furkat [3]
Racket : 94
balls : 12.50
warm ups : 34
shorts : 19.95
shoes : 44.50
total with tax : 204.95 + .05(204.95) = 204.95 + 10.25 = 215.20

20 per lesson for 12 lessons : 20 * 12 = 240
12 miles 1 way....thats 24 miles round trip, at 0.42 per mile : 24(0.42) = 
10.08.....lessons and mileage : 240 + 10.08 = 250.08

for a total of : 215.20 + 250.08 = 465.28 <==
4 0
3 years ago
Read 2 more answers
Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.
lakkis [162]

Answer:

2*sin(x)+y*cos(x)-cos(y)=C_1

Step-by-step explanation:

Let:

P(x,y)=2*cos(x)-y*sin(x)

Q(x,y)=cos(x)+sin(y)

This is an exact differential equation because:

\frac{\partial P(x,y)}{\partial y} =-sin(x)

\frac{\partial Q(x,y)}{\partial x}=-sin(x)

With this in mind let's define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x}=P(x,y)

and

\frac{\partial f(x,y)}{\partial y}=Q(x,y)

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y)

f(x,y)=\int\  2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}

Now, let's replace the previous result into \frac{\partial f(x,y)}{\partial y}=Q(x,y) :

cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)

Solving for \frac{dg(y)}{dy}

\frac{dg(y)}{dy}=sin(y)

Integrating both sides with respect to y:

g(y)=\int\ sin(y)  \, dy =-cos(y)

Replacing this result into f(x,y)

f(x,y)=2*sin(x)+y*cos(x)-cos(y)

Finally the solution is f(x,y)=C1 :

2*sin(x)+y*cos(x)-cos(y)=C_1

7 0
3 years ago
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