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nignag [31]
3 years ago
9

How do u write 4408730 in expanded form?

Mathematics
1 answer:
STALIN [3.7K]3 years ago
5 0
Do you need the expanded notation form, or expanded factor form?
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Amanda got 4/5 of the questions on the math quiz correct. Joe got 75% of the questions on the math quiz right. Who got more ques
Elden [556K]

Amanda got more questions correct.

<u>Step-by-step explanation:</u>

Given:

Amanda got  \frac{4}{5}  of the questions correct.

Joe got 75% or \frac{75}{100} of the questions correct.

To Find:

Individual with maximum number of correct question.

Solution:

Now we have to convert fraction and percent into a decimal number, so that we can compare it.

\frac{4}{5} = 0.8

\frac{75}{100} = 0.75,

Thus from the above obtained decimal details we can conclude that Amanda got more questions correct.

4 0
3 years ago
Evaluate the expression 12p2
scZoUnD [109]
I believe the answer 12p^2
7 0
3 years ago
Select the correct answer.
gtnhenbr [62]

Answer:

5 years

Step-by-step explanation:

To determine the number of years needed for the company to process a certain amount of milk, we use the given function which relates the number of years, x, and the gallons of milk, y. We simply substitute the number of gallons to y and solve for x.  

y = x^6 − 9x^4 + 70x^2

630 = x^6 − 9x^4 + 70x^2

Solving for x, we will have

x = 4.8866 years or approximately 5 years

Hope this answer helps you :)

Have a great day

Mark brainliest

7 0
3 years ago
The square below represents one whole.<br> What percent is represented by the shaded area?<br> %
sertanlavr [38]

Answer:

77%

Step-by-step explanation:

5 0
3 years ago
y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the
vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

For the second condition, we need to find the derivative y' first. In this case, we have:

y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.

Therefore:

y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.

This means that we must solve the following system of equations:

\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.

If we add the equations above, we get:

\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e}  \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.

If we now substitute c_1 = 0 into either of the equations in the system, we get:

c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}

This means that the solution obeying the initial conditions is:

\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

Indeed, we can see that:

y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3

y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,

which do correspond to the desired initial conditions.

3 0
3 years ago
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