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maxonik [38]
3 years ago
13

Caroline is making macaroni and cheese. She needs 3 cups of cheese to make 6 servings. How much cheese will Caroline need to mak

e two servings?
Mathematics
2 answers:
aleksandrvk [35]3 years ago
7 0
She needs one cup of cheese for two servings.
Anastaziya [24]3 years ago
3 0

Answer:1/2 of a cup.

Step-by-step explanation:

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What is the GCF of 100 and 20
alexdok [17]

Answer:

20

Step-by-step explanation:

20 is the GCF or greatest common factor of 20 and 100.

4 0
3 years ago
I need help with this problem
Anarel [89]

The midpoint lines are half the opposite side length.

Line DE is given as 7 , So Line F = 7 x 2 = 14

Line EF is given as 12, so Line D = 12 x 2 = 24

Line DF is given as 16, so line E = 16 x 2 = 32


The perimeter is the sum of the 3 sides:

Perimeter = 14 + 24 + 32 = 70

6 0
4 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
A cubic function with roots at 2, 0 and 3 containing the point at (5,-6)
natulia [17]

Answer:

Cubic equations and the nature of their roots

Just as a quadratic equation may have two real roots, so a cubic equation has possibly three. But unlike a quadratic equation which may have no real solution, a cubic equation always has at least one real root.

5 0
3 years ago
Evaluate (-1)(-1)(-3): <br> 3 <br> -3 <br> 5 <br> -5
Neporo4naja [7]

Answer:

-3

Step-by-step explanation:

Note that when you multiply two negative numbers, the answer will result as a positive number. Also note that when you multiply a negative number with a positive number, then your answer will result in a negative number.

(-1)(-1) = 1

(1)(-3) = -3

-3 is your answer.

~

5 0
3 years ago
Read 2 more answers
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