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3 years ago

Fill in the blanks to the picture above

Mathematics

2 answers:

Lena [83]3 years ago

5 0

Answer:

The answer is B, 6x^2 - 21x + 15

Step-by-step explanation:

= (3x + − 3) (2x + − 5)

= (3x) (2x) + (3x) (− 5) + ( − 3) (2x) + ( − 3) ( − 5)

= 6x^2 − 15x − 6x + 15

= 6x^2 − 21x + 15

- I don't know how you want me to fill the boxes exactly but here's the step by step explanation. I hope I helped.

Savatey [412]3 years ago

4 0

Middle one is the answer

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What is the line slope of (10,5) and (1,9)

sveta [45]

Use the slope formula
y2-y1
m=---------
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(rise-->y over run-->x)
so it would be
9-5
m=-------
1-10
and that would simplify to
4
----
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and that would be your slope.

Hope that helped :-)

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3 years ago

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What is the length of the strip of seaweed around the outside of the sushi?

SashulF [63]

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3 years ago

Which function has an asymptote at x= 39

LekaFEV [45]

Answer:

option B : $f(x)= log_7(x-39)$

Step-by-step explanation:

(a) $f(x) = 7^{x-39}$

For exponential function , there is no vertical asymptotes

General form of exponential function is

$f(x) = n^{ax-b}+k$

$f(x) = 7^{x-39}$

In the given f(x) the value of k =0

So horizontal asymptote is y=0

(b) lets check with option

$f(x)= log_7(x-39)$

To find vertical asymptote we set the argument of log =0 and solve for x

Argument of log is x-39

x-39=0 so x=39

Hence vertical asymptote at x=39

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3 years ago

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Fed [463]

the answer would be D.

Hope this helps! :)

~Zane

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3 years ago

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