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Katen [24]
3 years ago
14

8) Write an absolute value equation that has 5 and 15 as its solutions?

Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0
(I'm going to use brackets as my absolute value bars lol)
 [5 x -3]
[-15]
=15
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Can you guys please help me? Thanks!
olga2289 [7]
you are correct, it is one to two
4 0
3 years ago
Use conversion factors to find the answer, distance tracked in feet after 12 seconds at 87 miles per hour.
Assoli18 [71]
1 hour=3600 seconds
1 mile=5280 feet
87 miles/hr (1 hr/3600 sec)(5280 ft/mile)=127.6 feet/sec

Distance= Rate*time
d=127.6(12)=1531.2 feet (after 12 seconds) 
3 0
3 years ago
Help ? Plz forgot the steps (3z+9)(z-4)
Dafna11 [192]
It's a binomial times a binomial, so you FOIL
First multiply the First numbers
3z*z = 3z(^2)
Then multiply Outers:
3z* -4  = -12z 
Nest, Inners:
9*z = 9z
Finally, the Lasts: 
-4*9 = -36
Then, throw it all together in descending order, starting with the variable of highest degree to lowest:
3z^2 - 12z + 9z - 36
Last of all, simplify coefficients with like terms and you're good to go:
3z^2 - 3z - 36
Hope this helps!
3 0
3 years ago
100 PNTS AND BRAINLIST FOR CORRECT ANWSER WITH A GOOD EXPLAINATION!!
Lelechka [254]

Answer:

Expected value is -3

You should not play

Step-by-step explanation:

five are red, five are green, 10 yellow

P(red) = red/total = 5/20 = 1/4

P(green) = green/total = 5/20 = 1/4

P(yellow) = yellow/total =10/20 = 1/2

Expected value = 1/4(18) + 1/4(10) + 1/2(0) - 10 to play

                             =18/4 + 10/4 - 10

                            =28/4 -10

                             =7-10

                             = -3

You should not play

5 0
2 years ago
What's the correct answer for this?
11Alexandr11 [23.1K]
\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad  y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\

\bf \textit{\underline{N} is inversely proportional to \underline{A}}\qquad N=\cfrac{k}{A}
\\\\\\
\textit{we also know that }
\begin{cases}
A=3\\
N=16
\end{cases}\implies 16=\cfrac{k}{3}\implies 16\cdot 3=k
\\\\\\
48=k\qquad thus\qquad \boxed{N=\cfrac{48}{A}}\\\\
-------------------------------\\\\
\textit{if \underline{A} is 4, what is \underline{N}?}\qquad N=\cfrac{48}{4}
4 0
3 years ago
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