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Andrej [43]
3 years ago
8

Given A(-3,6) and B(5,-6) find the distance and midpoint between each

Mathematics
2 answers:
bezimeni [28]3 years ago
8 0
AB=\sqrt{(x _{b}- x_{a})^2+(y _{b}-y _{a})^2    }=  \sqrt{(5-(-3))^2+(-6-6)^2}\\
 \sqrt{8^2+(-12)^2}= \sqrt{64+144} = \sqrt{208}=4 \sqrt{13}
xenn [34]3 years ago
7 0
14.42 is your distance formula 
(1,0) is the midpoint 
if you need the forma for distance it is x2-x1 squared + y2-y1 squared then square root it 
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Jon sold 3/4 of his books and purchased 3 more. If Jon has 11 books now how many books did he have to begin with?
slava [35]
I hope this helps you



begin jons all book 4a


sold 4a.3/4=3a


remaining book 4a-3a=a

3 purcased book a+3=11

a=8


4a=4.8=32 jons all book
5 0
3 years ago
Evaluate the following expression. 7 × 22 + (8 - 2) ÷ 3
sweet-ann [11.9K]

Answer:

156

Step-by-step explanation:

Using PEMDAS: We do the paranthese first:

7 * 22 + (8-2) / 3.

7 * 22 + 6 / 3.

Multiply / Divide:

154 + 6 / 3

154 + 2

156

5 0
3 years ago
Read 2 more answers
The height of a tree increases at the rate of 1 cm per month. Remy begins measuring the tree at 10 cm. Identify a linear equatio
Mademuasel [1]

Answer:

18 cm tall

Step-by-step explanation:

we start with 10

then each month we add 1 over 8 months

so  10+8=18

3 0
3 years ago
Factor the Polynomial: 51c^3-34c
labwork [276]

Answer:

17 c (3 c^2 - 2)

Step-by-step explanation:

Factor the following:

51 c^3 - 34 c

Hint: | Factor common terms out of 51 c^3 - 34 c.

Factor 17 c out of 51 c^3 - 34 c:

Answer: 17 c (3 c^2 - 2)

3 0
3 years ago
The line containes the point (-8,11) and has a y intercept of 5
kicyunya [14]

Step-by-step explanation:

Since, line contains the point (-8, 11) and has a y intercept of 5.

Therefore, line passes through the points (-8, 11) and (0, 5)

Equation of line in two point form is given as:

\frac{y-y_1 }{y_1 - y_2 }  =  \frac{x-x_1 }{x_1 - x_2 }  \\  \\  \therefore \:  \frac{y-11 }{11 - 5 }  =  \frac{x-( - 11) }{ - 11 - 0 }  \\  \\ \therefore \:  \frac{y-11 }{6 }  =  \frac{x +  11 }{ - 11 }  \\  \\ \therefore \:   - 11(y-11 )=6 (x +  11) \\  \\  \therefore \:   - 11y + 121=6 x +  66 \\  \\  \therefore \:   0=6 x + 11y +  66 - 121 \\  \\    \red{ \boxed{\therefore \:  6 x + 11y  - 55 = 0}}

Hence, 6x + 11y - 55 = 0 is the required equation of line.

5 0
3 years ago
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