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Tpy6a [65]
3 years ago
11

Complete the square to rewrite y = x– 5x + 7 in graphing form and state the vertex.

Mathematics
1 answer:
Sati [7]3 years ago
3 0

Answer:

Slope = -4

Y-intercept= (0,7)

Step-by-step explanation:

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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]

Yes. Conceptually, all the matrices in the group have the same structure, except for the variable component x. So, each matrix is identified by its top-right coefficient, since the other three entries remain constant.

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]

is an isomorphism.

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]

is clearly the image of the real number x.

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)

8 0
3 years ago
If $20.00 is for 12.25kg of peas ...what weight of peas can i get for $4.00​
Alina [70]

2.45kg

First you calculate the price per kg as follows:

If it’s $20 for 12.25 kg, the price per kg is $20/12.25=$1.63 per kg


If you have $4, you can figure out what weight you can get by taking $4/the price per kg, so $4/$1.63, which is 2.45 kg.

4 0
3 years ago
Explain please thanks :)
dem82 [27]

Answer:

24

Step-by-step explanation:

The question is saying, how many three digit numbers can be made from the digits 3, 4, 6, and 7 but there can't be two of the same digit in them. For example 346 fits the requirements, but 776 doesn't, because it has two 7s.

Okay, on to the problem:

We can do one digit at a time.

First digit:

There are 4 digits that we can choose from. (3, 4, 6, and 7)

Second digit:

No matter which digit we chose for the first digit, there is only going to be 3 of them left, because we already chose one, and you can't repeat that same digit. So there are 3 options.

Third digit:

Using the same logic, there are only 2 options left.

We have 4 choices for the first digit, 3 choices for the second, and 2 for the third.

Hence, this is 4 * 3 * 2 = 24 three-digit numbers that can be made.

8 0
2 years ago
Translate this sentence into an equation. 128 is the product of Donnie's age and 8 . Use the variable d to represent Donnie's ag
Alina [70]
128 = 8d, because the product calls for multiplication to be performed.
8 0
3 years ago
Read 2 more answers
in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference​
allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}

so that

a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d

a_n=a_1+(n-1)d

33=8+(n-1)d

21=(n-1)d

n has to be an integer, which means there are 4 possible cases.

Case 1: n-1=1 and d=21. But

\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123

Case 2: n-1=21 and d=1. But

\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123

Case 3: n-1=3 and d=7. But

\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123

Case 4: n-1=7 and d=3. But

\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123

8 0
3 years ago
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