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aleksklad [387]

3 years ago

8

Gary had a pizza party for his friend 14 people showed up,and they had 50 pieces of pizza. How many pieces of pizza did each per

son have

2 answers:

andreev551 [17]3 years ago

8 0

Answer:

3

Hope this helps

Vilka [71]3 years ago

5 0

Answer:

3

Step-by-step explanation:

each friend would have 3 pizzas

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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

Help!!!! me please the first to answer will get brainliest!!!! please help me<br><br><br>

never [62]

<h3>1 Answer: v = -130</h3>

=============================================================

Explanation:

Because we're given multiple choices to pick from, we can go through each to do trial and error.

For instance, if v = 52, then

v/13 < -7

52/13 < -7

4 < -7

but that's a false statement since 4 is not smaller than -7. Use a number line to see this. You should have -7 to the left of 4, showing that -7 is the smaller value.

Because 4 < -7 is false, this means v/13 < -7 is false when v = 52. This rules out v = 52. You should find that v = 26 and v = -39 won't work for similar reasons.

Only v = -130 works since...

v/13 < -7

-130/13 < -7

-10 < -7

This is true because -10 is to the left of -7 on the number line.

-------------------------------------------------

Another approach:

We have some unknown number v, and we're dividing by 13 to get the expression v/13. To undo this, we multiply both sides by 13

v/13 < -7

13*(v/13) < 13*(-7) .... multiply both sides by 13

v < -91

So any solution to the original inequality must be smaller than -91. This allows us to rule out v = 52, v = 26 and v = -39.

The value v = -130 is smaller than -91, so it satisfies v < -91. Therefore, v = -130 is a solution to the original inequality.

6 0

2 years ago

Please help me I’m really stuck

dalvyx [7]

Answer:

<u>/</u><u> </u>3=<u>/</u><u> </u>5 { alternate anglea}

<u>/</u><u> </u>5+<u>/</u>6 =180° {straight angle}

<u>/</u><u> </u>6=180-26

<u>/</u><u> </u>6=154°

hope it helps

<h3>stay safe healthy and happy.</h3>

6 0

3 years ago

Spencer bought 5 1/2 pounds of potatoes and 3 3/4 pounds of tomatoes to make stew.

Advocard [28]

Answer:

9 1/4 pounds because you add the two fractions together

6 0

3 years ago

Read 2 more answers

Triangle ABC has vertices A(-2, 3), B(0, 3), and C(-1,-1). Find the coordinates of the image after a reflection over the

motikmotik

<u>Given</u>:

Given that the triangle ABC has vertices A(-2,3), B(0,3) and C(-1,-1).

We need to determine the coordinates of the image after a reflection over the x - axis.

Let A'B'C' denote the coordinates of the triangle after a reflection over the x - axis.

<u>Coordinates of the point A':</u>

The general rule to reflect the coordinate across the x - axis is given by

$(x,y)\rightarrow (x,-y)$

Substituting the point A(-2,3), we get;

$(-2,3)\rightarrow (-2,-3)$

Thus, the coordinates of the point A' is (-2,-3)

<u>Coordinates of the point B':</u>

The general rule to reflect the coordinate across the x - axis is given by

$(x,y)\rightarrow (x,-y)$

Substituting the point B(0,3), we get;

$(0,3)\rightarrow (0,-3)$

Thus, the coordinates of the point B' is (0,-3)

<u>Coordinates of the point C':</u>

The general rule to reflect the coordinate across the x - axis is given by

$(x,y)\rightarrow (x,-y)$

Substituting the point C(-1,-1), we get;

$(-1,-1)\rightarrow (-1,1)$

Thus, the coordinates of the point C' is (-1,1)

Hence, the coordinates of the image after a reflection over the x - axis is A'(-2,-3), B(0,-3) and C(-1,1)

8 0

3 years ago