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e-lub [12.9K]
1 year ago
7

The points (3,-10) and (7,r) lie on a line with slope 5/4. Find the missing coordinate r.

Mathematics
1 answer:
LenaWriter [7]1 year ago
5 0

(\stackrel{x_1}{3}~,~\stackrel{y_1}{-10})\qquad (\stackrel{x_2}{7}~,~\stackrel{y_2}{r}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{r}-\stackrel{y1}{(-10)}}}{\underset{run} {\underset{x_2}{7}-\underset{x_1}{3}}} \implies \cfrac{r +10}{4}~~ = ~~\stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{4}} \\\\\\ 4r+40=20\implies 4r=-20\implies r=\cfrac{-20}{4}\implies {\Large \begin{array}{llll} r=-5 \end{array}}

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Find the length of the arc intercepted by an angle of 36° in a circle with radius 12 inches
azamat
Set up a pair of proportional ratios...

a/(2πr)=α/360

a=(πrα)/180, we are given that r=12 and α=36° so

a=432π/180 in

a=12π/5 in

a=2.4π in

a≈7.54 in (to nearest one-hundredth of an inch)



8 0
3 years ago
Rewrite to make true: There are five terms in the series (attached)
Blababa [14]

Answer:

There are six terms in the series

Step-by-step explanation:

We can see that n starts from 0, and ends at 5. There are therefore 6 terms in this series - as 0 is included. Therefore there will be 6 terms in this series, not 5.

There are six terms in the given series.

4 0
3 years ago
Hey guys, i wanted to ask if both questions are correct ?
stira [4]
Yay you are right for both Question
8 0
3 years ago
2. How many 4-digit even numbers greater then 4,999 can be made(using the digits 0-9?
Lostsunrise [7]

Answer:

For each choice of the first two digits you have 10 choices for the third digit. Thus you have 10x10x10 = 1000 choices for the first three digits. Finally you have 10 choices for the fourth digit and thus there are 10x10x10x10 = 10 000 possible 4 digit combinations from 0-9.

I hope that helps! :)

3 0
3 years ago
Write the equation of the function whose graph is shown. y = (x + )2 +
Deffense [45]
Quadratic form is given in the form
y= ax^{2}+bx+c, where a, b, and c are constants

We substitute the value (8, 12) and (5,3) in turn

12= a(8)^{2}+b(8)+c
12=64a+8b+c ... Equation 1

3=a(5)^{2}+b(5)+c
3=25a+5b+c ... Equation 2

Equation 1 - Equation 2 gives
39a+3b=9, we call this Equation 3

We also know that (5, 3) is the turning point of the curve, where the value of x is given by x=- \frac{b}{2a}
Substitute x=5
5=- \frac{b}{2a}
10a=-b
b=-10a ... Equation 4

Substitute Equation 4 into Equation 3
39a+3b=9
39a+3(-10a)=9
39a-30a=9
9a=9
a=1

Now we need to work out the value of b and c

Substitute a=1 back into equation 4
-10a=b
-10(1)=b
b=-10

Substitute a=1 and b=-10 into either equation 1 or equation 2

64a+8b+c=12
64(1)+8(-10)+c=12
64-80+c=12
c=12+80-64
c=28

Hence the equation of the quadratic curve is
y= x^{2} 10x+28


8 0
3 years ago
Read 2 more answers
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