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1 year ago

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

1 answer:

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

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

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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.

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3 years ago

Hey guys, i wanted to ask if both questions are correct ?

stira [4]

Yay you are right for both Question

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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

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