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Arada [10]
2 years ago
15

Brianna has $50 in her bank account. She pays $6 to attend her school football game. She wants to buy hotdogs for her and her fr

iends. If each hotdog costs $2, how many hotdogs can she buy if she wants to maintain a balance of at least $20 left in her bank account? Write the inequality and solve it.
Mathematics
1 answer:
sattari [20]2 years ago
6 0

Answer:

12

Step-by-step explanation:

$50-6 (ticket to game) = 44 - $20 (the balance she wants to have left) = $24.

$24/2=12 (amount of hotdogs she can buy)

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-6x -(3x)= what's the answer to this please help me .
sleet_krkn [62]

-6x -(3x)

Pretend that there is a -1 in front of the bracket

-6x-1(3x)

mutiply the bracket by -1

(-1)(3x)= -3x

-6x-3x= -9x

Answer: -9x

3 0
3 years ago
Read 2 more answers
I don't even know how to figure this out. I don't have a formula for this
yuradex [85]

Answer:

\sin\theta=\dfrac{15}{17}

?=15

Step-by-step explanation:

Use the trigonometry formula

\cos^2\theta+\sin^2\theta=1

You are given that

\cos \theta=\dfrac{8}{17}

Substitute into the first formula

\left(\dfrac{8}{17}\right)^2+\sin^2\theta=1\\ \\\dfrac{64}{289}+\sin^2\theta=1\\ \\\sin^2\theta=1-\dfrac{64}{289}=\dfrac{289-64}{289}=\dfrac{225}{289}\\ \\\sin \theta=\pm \dfrac{15}{17}

Angle \theta is acute angle, then the sine of this angle is positive and

\sin\theta=\dfrac{15}{17}

3 0
3 years ago
Car = V-8 sedan Years driven = first through third Miles driven = 14,000 miles each year In dollars and cents, the total project
saveliy_v [14]

Answer:

609

Step-by-step explanation:

5 0
3 years ago
The distance from Earth to the moon is 384,400 kilometers. What is this distance expressed in scientific notation?
umka2103 [35]
3.844 times 106 kilometers
5 0
3 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
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