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nalin [4]
3 years ago
7

Whoever needed help with this MATH question I got it for you... Answer: F(x)= 4,800(1.5)^X

Mathematics
2 answers:
Ganezh [65]3 years ago
8 0

Answer:

1+2=3

Step-by-step explanation:

Bogdan [553]3 years ago
6 0

Answer:

thx brainlist pls

Step-by-step explanation:

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Please help figure out I will give brainliest!
noname [10]

Answer:

40 student 110 adult

this is using desmos

but to smash the equation you can go like basically we have to cancel out each variable to solve for the other

3x + 8y = 1000

x + y < =150

-3(x + y <= 150)

-3x -3y = -450

3x + 8y = 1000

5y = 550

y= 110

3x + 8 y = 1000

-8(x+y<=150) = -8x - 8y <= -1200

-5x = -200

\frac{ - 5x}{ - 5 }   = x

\frac{ - 200}{ - 5}  = 40

x= 40

40 kids 110 adults

7 0
2 years ago
Is my answer correct
Vlad [161]
No because its explain how much cumulative tip total on day 5 so the answer is D.400
8 0
3 years ago
Need help finding a domain
CaHeK987 [17]

Answer:

the domains is

- 10 < x \leqslant 10

7 0
2 years ago
Find the hypotenuse of an isosceles right triangle when the legs each measure 3\sqrt{x} 2inches.
Tresset [83]

The hypotenuse of the right triangle measures 3 units.

<h3>How to find the hypotenuse?</h3>

Here we know that both legs measure:

L  =\frac{3}{\sqrt{2} }

If we use the Pythagorean theorem, we will see that the hypotenuse H can be written as:

H^2 = (\frac{3}{\sqrt{2} } )^2 +  (\frac{3}{\sqrt{2} } )^2 \\\\H^2 = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9\\\\H = \sqrt{9} = 3

Then we conclude that the hypotenuse of the right triangle measures 9 units.

If you want to learn more about right triangles:

brainly.com/question/2217700

#SPJ1

6 0
2 years ago
Plz help me!!!!!!!!!
Over [174]

Answer:  \bold{\dfrac{4\pm \sqrt{6}}{2}}

<u>Step-by-step explanation:</u>

\dfrac{3}{y-2}-2=\dfrac{1}{y-1}\\\\\\\text{Multiply by the LCD (y-2)(y-1) to clear the denominator:}\\\\\dfrac{3}{y-2}(y-2)(y-1)-2(y-2)(y-1)=\dfrac{1}{y-1}(y-2)(y-1)\\\\\\3(y-1)-2(y-2)(y-1)=1(y-2)\\\\3y-3-2(y^2-3y+2)=y-2\\\\3y-3-2y^2+6y-4=y-2\\\\-2y^2+9y-7=y-2\\\\0=2y^2-8y+5\quad \rightarrow \quad a=2,\ b=-8,\ c=5\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(5)}}{2(2)}\\\\\\.\ =\dfrac{8\pm \sqrt{64-40}}{2(2)}

.\ =\dfrac{8\pm \sqrt{24}}{2(2)}\\\\\\.\ =\dfrac{8\pm 2\sqrt{6}}{2(2)}\\\\\\.\ =\dfrac{4\pm \sqrt{6}}{2}

7 0
3 years ago
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