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3 years ago

9

Donovan wants to earn a 70% in Mr. Mottola's class. He knows he currently has an 67% in the class after 9 grades. What is the mi

nimum grade Donovan would need to earn in order to have a 70%.

1 answer:

Likurg_2 [28]3 years ago

6 0

Donovan needs to earn at least a 73% in order to have a 70%

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Solve by the addition method

Sladkaya [172]

Answer is no solution I believe

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3 years ago

Ms. Forte’s class is studying plants. 3 students want to share a packet of 14 seeds equally. What is each child’s share?

vekshin1

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3 years ago

Read 2 more answers

4. The red graph (1) is the graph of f(x) = log(x). Describe the transformation of the blue function (2) and write the equation

sergey [27]

Answer:

Function $f(x)$ is shifted 1 unit left and 1 unit up.

$f(x)\rightarrow f(x+1)+1$

Transformed function $f(x+1)+1=\log(x+1)+1$

Step-by-step explanation:

Given:

Red graph (Parent function):

$f(x)=\log(x)$

Blue graph (Transformed function)

From the graph we can see that the red graph is shifted 1 units left and 1 units up.

Translation Rules:

$f(x)\rightarrow f(x+c)$

If $c>0$ the function shifts $c$ units to the left.

If $c$ the function shifts $c$ units to the right.

$f(x)\rightarrow f(x)+c$

If $c>0$ the function shifts $c$ units to the up.

If $c$ the function shifts $c$ units to the down.

Applying the rules to $f(x)$

The transformation statement is thus given by:

$f(x)\rightarrow f(x+1)+1$

As function $f(x)$ is shifted 1 unit left and 1 unit up.

Transformed function is given by:

$f(x+1)+1=\log(x+1)+1$

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3 years ago

Find the solution to the system of equations.

Anna [14]

No solution it’s infinite

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2 years ago

5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

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3 years ago