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Flauer [41]
4 years ago
14

Henry is following the recipe card to make a cake. He has 95 cups of flour. How many cakes can Henry make?

Mathematics
2 answers:
algol134 years ago
8 0

Answer:

30 cakes

Step-by-step explanation:

ahrayia [7]4 years ago
3 0
For one cake alone, find a common denominator (which is 6). So 2 2/3 (8/3) is equal to 16/6 cups. 1/2 is 3/6 cups. So for one cake, it takes 19/6 cups of flour. Henry has 570/6 (=95) cups of flour. If you divide the 570 by 19, Henry can make 30 cakes.
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Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
2 2-1-1-1 whats is the answer
schepotkina [342]
2-1=1-1=0-1=-1 is the first 2 part of the eqation too then it will be 2-2=0-1=-1-1=-2-1=-3
-3
3 0
3 years ago
Solve y′′=sin(x) if y(0)=0 and y′(0)=5.<br><br> y(x)=?
Ksenya-84 [330]

Answer:

y(x)=6x-sin(x)

Step-by-step explanation:

Rewrite the differential equation as:

\frac{d^{2} y }{dx^{2} } =sin(x)

Integrate both sides with respect to x:

\int\ \frac{d^{2} y }{dx^{2} } dx = \int\ sin(x) dx

\frac{dy}{dx} =-cos(x)+C_1

Integrate one more time both sides with respect to x:

\int\ \frac{dy}{dx} = \int\ -cos(x)+C_1 dx

y(x)=-sin(x)+C_1x+C_2

Now that we find the solution, let's find its derivate:

y'(x)=C_1-cos(x)

Evaluating the initial conditions:

y(0)=C_1(0)+C_2-sin(0)=0\\C_2=0

y'(0)=C_1-cos(0)=5\\C_1=5+1=6

Replacing the value of the constants that we found in the differential equation solution:

y(x)=6x-sin(x)

6 0
3 years ago
PLEASE HELP I WILL MARK BRAINLIEST FOR THE ANSWERS! PLEASE HELP PLEASE ITS DUE AT 12
sesenic [268]

Answer:

61

Step-by-step explanation:

84+38=122

180-122=58

DBC=58

180-58=122

122÷2=61

61+58=119

180-119=61

3 0
2 years ago
Read 2 more answers
6x5x7xfreepoints5x7x8x6x45x6x8x7x55x6x6x6x6x0=
Liula [17]

Answer: 0

Yurrrrrrr

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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