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Maksim231197 [3]
2 years ago
6

Graph the line with the equation y = 1/6x-5 PLEASE HELP

Mathematics
1 answer:
quester [9]2 years ago
5 0

Answer:

Step-by-step explanation:

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Simplify (1-6)(-4-7)​
lakkis [162]

Answer:

I’m guessing you subtract the numbers from the inside of the parenthesis and then multiply the two.

(-5)(-11) = 55

I hope this is correct.

7 0
3 years ago
What is the measure of angle 1? Put the number only.
Katarina [22]

Answer:

angle 1 = 120 degree

Step-by-step explanation:

75 degree + 45 degree = angle 1 (sum of two opposite interior angles is equal to the exterior angle formed)

120 degree = angle 1

6 0
2 years ago
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What is log6^e f rewritten using the power property?
marishachu [46]

Given expression : log_6e^f ..

We need to apply power property of logs to rewrite it.

According to log rule of exponents:

log_b a^n = n log_b(a)

If we compare given expression with the rule the exponent part is f, base is 6.

Therefore, we need to bring exponent f in front of log.

Therefore, log_6e^f  = flog_6 e.

<h3>And correct option is second option flog_6 e.</h3>
4 0
3 years ago
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Enter the range of data 37,45,20,34,28,80
Montano1993 [528]

Answer:

60

Step-by-step explanation:

To find the range, you subtract the largest number and the smallest number.

Largest number: 80

Smallest number: 20

80-20=60

8 0
3 years ago
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Determine whether the equation is exact. If it​ is, then solve it. e Superscript t Baseline (7 y minus 3 t )dt plus (2 plus 7 e
umka21 [38]

Answer:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

Step-by-step explanation:

You have the following differential equation:

e^t(7y-3t)dt+(2+7e^t)dy=0

This equation can be written as:

Mdt+Ndy=0

where

M=e^t(7y-3t)\\\\N=(2+7e^t)

If the differential equation is exact, it is necessary the following:

\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}

Then, you evaluate the partial derivatives:

\frac{\partial M}{\partial y}=\frac{\partial}{\partial t}e^t(7y-3t)\\\\\frac{\partial M}{\partial t}=7e^t\\\\\frac{\partial N}{\partial t}=\frac{\partial}{\partial t}(2+7e^t)\\\\\frac{\partial N}{\partial t}=7e^t\\\\\frac{\partial M}{\partial t} = \frac{\partial N}{\partial t}

The partial derivatives are equal, then, the differential equation is exact.

In order to obtain the solution of the equation you first integrate M or N:

F(t,y)=\int N \partial y = (2 +7e^t)y+g(t)        (1)

Next, you derive the last equation respect to t:

\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)

however, the last derivative must be equal to M. From there you can calculate g(t):

\frac{\partial F(t,y)}{\partial t}=M=(7y-3t)e^t=7ye^t+g'(t)\\\\g'(t)=-3te^t\\\\g(t)=-3\int te^tdt=-3[te^t-\int e^tdt]=-3[te^t-e^t]

Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

where C is the constant of integration

8 0
3 years ago
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