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Lemur [1.5K]
3 years ago
5

John wants to make cookies. To make cookies, he need 2 over 3 of a cup of flour per batch of cookies. If John has 4 cups of flou

r then how many batch of cookies can John make?
Mathematics
1 answer:
Mnenie [13.5K]3 years ago
6 0
\dfrac{2}{3} \text { cups = } 1 \text { batch}

----------------------------------------------
Find 1 cup :
----------------------------------------------

1 \text { cup = } 1 \div  \dfrac{2}{3} \text { batch}

1 \text { cup = } 1 \times  \dfrac{3}{2} \text { batch}

1 \text { cup = } \dfrac{3}{2} \text { batch}

----------------------------------------------
Find 4 cups :
----------------------------------------------
4 \text { cups = } \dfrac{3}{2} \times 4 \text { batch}

4 \text { cups = } \6 \text { batches}

--------------------------------------------------------------------------------------------
Answer: John can make 6 batches with 4 cups of flour.
--------------------------------------------------------------------------------------------
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3 years ago
z varies directly with x and inversely with y^2. When x = 2 and y = 5, z = 8. What is the value of z when x = 4 and y = 9?
Ivanshal [37]
Direct variation is  y = kx

Inverse variation is y = k/y

if z varies directly with x, then the equation would be z = kx

if z varies inversely with y^2 then the equation would be z = k/y^2. Combining, these two, we have z = kx/y^2

We use the first set of values to find the constant k:

x = 2, y = 5, z = 8:

8 = 2x/25, k = 100. Now we use the second set of values to find z:

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4 0
3 years ago
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Let g' be the group of real matricies of the form [1 x 0 1]. Is the map that sends x to this matrix an isomorphism?
aliina [53]

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However, let's prove in a more formal way that

\phi:\ \mathbb{R} \to G,\quad \phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]

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First of all, it is injective: suppose x \neq y. Then, you trivially have \phi(x) \neq \phi(y), because they are two different matrices:

\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right],\quad \phi(y) = \left[\begin{array}{cc}1&y\\0&1\end{array}\right]

Secondly, it is trivially surjective: the matrix

\phi(x) = \left[\begin{array}{cc}1&x\\0&1\end{array}\right]

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Finally, \phi and its inverse are both homomorphisms: if we consider the usual product between matrices to be the operation for the group G and the real numbers to be an additive group, we have

\phi (x+y) = \left[\begin{array}{cc}1&x+y\\0&1\end{array}\right] = \left[\begin{array}{cc}1&x\\0&1\end{array}\right] \cdot \left[\begin{array}{cc}1&y\\0&1\end{array}\right] = \phi(x) \cdot \phi(y)

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Answer:

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