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Hoochie [10]
3 years ago
9

Jennie has 7 tickets for carnival rides. The Ferris wheel costs 4 tickets, The roller coaster costs 6 tickets. the merry-go-roun

d cost 3 tickets. Jennie wants to ride all three rides. She correctly writes and solves an equation to find t, the number of additional tickets she must buy. Which could show Jennie's work?
Mathematics
1 answer:
algol133 years ago
6 0

Answer: 6

Step-by-step explanation:

7 - 3 = 4

3 - 4 = 0

0 - 6 = -6

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How do you do this, I don’t understand it
Otrada [13]
For the first line start at -1 and from there go up to and to the right three
for the second line on nuber one start at 4 and down 1 and to the left 1
7 0
3 years ago
Read 2 more answers
Write down the output y in terms of x
atroni [7]

Answer:

the x means 9

Step-by-step explanation:

6 0
3 years ago
On a trip to his grandparents, Donny drove 60 miles per hour for 45 minutes and 50 miles per hour for 30 minutes. How many miles
LenaWriter [7]

Donny drove 70 miles.

Since 60 miles per hour is 1 mile a minute so for in 45 minutes he drove 45 miles. After he drove for 50 miles for thirty minutes so in totally he drove 25 since 50 divided by two is 25.

7 0
3 years ago
How to show these 2 problems are inverses
nataly862011 [7]
\bf \begin{cases}
f(x)=\sqrt[3]{7x-2}\\\\
g(x)=\cfrac{x^3+2}{7}
\end{cases}\\\\
-----------------------------\\\\
now
\\\\
f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2}
\\\\\\
\sqrt[3]{x^3}\implies x\\\\
-----------------------------\\\\
or
\\\\
g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7}
\\\\\\
\cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x

thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other
8 0
3 years ago
Please show me how to solve the initial value problem <br> y'=tanx y(pi/4)=3
AlexFokin [52]
The ODE is separable, i.e. you can write

\dfrac{\mathrm dy}{\mathrm dx}=\tan x\iff\mathrm dy=\tan x\,\mathrm dx

Integrating both sides gives the general solution.

\displaystyle\int\mathrm dy=\int\tan x\,\mathrm dx
y=-\ln|\cos x|+C

Given that y\left(\dfrac\pi4\right)=3, we have

3=-\ln\left|\cos\dfrac\pi4\right|+C
3=-\ln\dfrac1{\sqrt2}+C
3-\ln\sqrt2=C


and so the particular solution to the IVP is

y=-\ln|\cos x|+3-\ln\sqrt2
y=3-\ln|\sqrt2\cos x|
4 0
4 years ago
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