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borishaifa [10]
4 years ago
6

Solve the system of equations using addition.

Mathematics
1 answer:
lutik1710 [3]4 years ago
4 0

Answer:

A. (-3,-6)

Step-by-step explanation:

The explanation is in the pictures.

You might be interested in
The question is the photo
Readme [11.4K]
For this question you can say:
5/9( °F - 32) = -10 °C
so now you should solve the equation:
5F/9 - 160/9 = -10 
⇒ 5F/9 = -10 + 160/9 ⇒ 5F/9 = -90/9 + 160/9 ⇒ 5F/9 = 70/9 ⇒ now you can multiply both sides of the equation by 9 so you will have:
5F = 70 ⇒ F = 14 ° F :)))
i hope this is helpful
have a nice day 
6 0
3 years ago
Which fraction is between 0 and 1/2
Greeley [361]
A fraction between 0 and 1/2

What is smaller than 1/2?

Smaller than 0.5 is 0.3

0.3 = 1/3

A fraction smaller than 1/2 and bigger than 0 is 1/3

Final answer:        1/3 
5 0
4 years ago
Mark is reading a book with 96 pages. It takes him 4 days to read the book. He reads the same number of pages each day. How many
gogolik [260]

24

Step-by-step explanation:

he reads 24 pages a day

7 0
3 years ago
Read 2 more answers
Dominic, Ralf and Peter have just eaten a HUGE Christmas meal. Dominic now weighs half as much as Ralf, and Peter weighs three t
Strike441 [17]
Dominic weighs 120.

I started laughing as soon as I saw this question...

It honestly made my day:)


4 0
3 years ago
A slope field produces...
balandron [24]

It's A. B doesn't really make sense ("derivative of the differential equation" is somewhat nonsensical, "derivative of an equation" is not meaningful).

More to the point: Slope fields are used to visualize solutions to differential equations of the form

<em>y'</em> = <em>f(x</em>, <em>y)</em>

You take some point (<em>x</em>, <em>y</em>) and evaluate <em>y'</em> at the point. This gives the slope of the line tangent to the particular solution to the DE that passes through the point (<em>x</em>, <em>y </em>).

Sample several points and evaluate <em>y'</em> at those points and you get several different slopes.

Simple example:

<em>y'</em> = <em>x</em> ² - <em>y</em> ² = (<em>x</em> + <em>y </em>) (<em>x</em> - <em>y</em> )

Let's take the points (1, 1), (-1, 0), and (2, -2), at which we get slopes

<em>y'</em> = <em>f</em> (1, 1) = 0

<em>y'</em> = <em>f</em> (-1, 0) = 1

<em>y'</em> = <em>f</em> (2, -2) = 0

From here, you can get particular solutions that pass through a certain point by interpolating the slopes of the tangents to the solution. I've attached a slope field for the example here at the points listed above. (In the order red, green, black). Each light gray arrow in the background shows the slope of the tangent line.

If you're still unsure how the slope field is generated, I suggest looking up videos on the subject. The process is a bit difficult to describe without a dynamic visual aid.

3 0
3 years ago
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