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3 years ago

Quick Question!

Mathematics

2 answers:

masya89 [10]3 years ago

8 0

It is unidentifiable, because nothing can be divided by 0

Artemon [7]3 years ago

7 0

Answer:

Undefined

Step-by-step explanation:

13×27÷6+8-17÷0×45×67×0×334 would equal undefined because you are dividing by 0. Anything divided by 0 is undefined.

If this answer is correct, please make me Brainliest!

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Scilla [17]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A pattern of ordered pairs is shown.

Anna007 [38]

Answer:

A. is the answer

Step-by-step explanation:

The first numbers are adding 2 and the second number is adding 3, hope this helped, plz brainiest.

8 0

2 years ago

Read 2 more answers

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago

A school P is 16km due west of a school Q. what is the bearing of Q from P

Irina-Kira [14]

Answer:

16Km due east of school P

Step-by-step explanation:

Given

A school P is 16km due west of a school Q

Thus, we can say that distance PQ = 16 km.

________________________________

now we have to find the bearing of Q from P

As distance is same

distance PQ = distance QP

Thus,

Distance will remain same of 16 km.

For direction,

If Q is west of P, then P will be east of Q $P------------------>Q$

as shown in figure P is west of Q,

now from point P , Q is west P.

Thus,

Bearing of School Q from P is 16Km due east of school P

7 0

3 years ago

Find f(-10) F(x) =7x-5 A= -75 B=-65 C=65 D=90

STALIN [3.7K]

Answer:

-75

Step-by-step explanation:

f(-10) = 7(-10) - 5

f(-10) = -70 - 5

f(-10) = -75

8 0

3 years ago