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arlik [135]
4 years ago
15

How do you find the distance with a number line ?

Mathematics
1 answer:
Marianna [84]4 years ago
3 0
The simplest way is to count each "line"

For example, #2
We start at J, which is at -5
Count the lines all the way to L, at 3
Distance would be 8
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Help !!!!!!!!!!!!!!!
Ostrovityanka [42]

Answer:

  x = 29

Step-by-step explanation:

Follow the directions: substitute 13 for y in the first equation.

  x + 13 = 42

Subtract 13 from both sides of the equation.

  x +13 -13 = 42 -13

  x = 29

4 0
3 years ago
Differentiate<br>y=tan (x^2-5x+6)​
slavikrds [6]

Use the chain rule:

<em>y</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)

<em>y'</em> = sec²(<em>x</em> ² - 5<em>x</em> + 6) × (<em>x</em> ² - 5<em>x</em> + 6)'

<em>y'</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)

Perhaps more explicitly: let <em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6, so that

<em>y(x)</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)   →   <em>y(u(x))</em> = tan(<em>u(x)</em> )

By the chain rule,

<em>y'(x)</em> = <em>y'(u(x))</em> × <em>u'(x)</em>

and we have

<em>y(u)</em> = tan(<em>u</em>)   →   <em>y'(u)</em> = sec²(<em>u</em>)

<em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6   →   <em>u'(x)</em> = 2<em>x</em> - 5

Then

<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>u</em>)

or

<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)

as we found earlier.

5 0
3 years ago
A triangle with an area of 45.5 cm2 has a base
FromTheMoon [43]

If you know the base and area of the triangle, you can divide the base by 2, then divide that by the area to find the height

Step 1. 14 ÷ 2 = 7

Step 2. 45.5 ÷ 7 = 6.5

Height = 6.5 cm

3 0
3 years ago
Y – 15 = 31<br>-2x + 5y = -3<br><br>A.(-6,3)<br><br>B.(-7,6)<br><br>C.(-3,6)<br><br>D.(-6,-3)​
Roman55 [17]

Answer:

D. ( -6, -3)

Step-by-step explanation:

believe 95% is correct

5 0
3 years ago
How do you write a b c d and e in an exponent way
sesenic [268]
Exponent way n
exponetional way
6 0
4 years ago
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