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Eva8 [605]
2 years ago
12

I don't know how to do this. Can someone explain how to please?​

Mathematics
1 answer:
Natalija [7]2 years ago
3 0

Answer:

f(5) = -2

Step-by-step explanation:

You look on the x-axis for 5 because f(5) means let x = 5. Then go down to the graph (the red zig-zaggy line) At the red line (you're still directly under the 5) look over to the y-axis to find the value. That's your answer. It is -2. The point (5, -2) is on the red graph.

f(5) = -2

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Don't answer if you don't know
SIZIF [17.4K]
0.10d + 0.25q = 3.55
q = d + 3

0.10d + 0.25(d + 3) = 3.55
0.10d + 0.25d + 0.75 = 3.55
0.10d + 0.25d = 3.55 - 0.75
0.35d = 2.80
d = 2.80/0.35
d = 8.....8 dimes

q = d + 3
q = 8 + 3
q = 11 <=== 11 quarters
3 0
3 years ago
The Ross family had 100 relatives at their big Thanksgiving dinner of all the guest nine he had French heritage 80 had English h
il63 [147K]
Hi, Deedee. I am quite confused with the numbers that you've given because they don't add up to 100. Nevertheless, if by 'this', you mean the Thanksgiving culture, then you would just count the English and the Native American heritage. This is because the first Thanksgiving dinner was shared between the English colonists and the Native American tribes.

I hope I was able to help you in a way. Have a good day.
8 0
3 years ago
The set of types of matter<br>well difined or not​
Tresset [83]

Answer:

It is not well defined.

Step-by-step explanation:

Because set is a collection of well defined objet. And matter contains of solid, liquid and gas

3 0
3 years ago
Freeeeeeeeeeeeeeeeee again​
Paraphin [41]

Answer:

cool

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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