Ask question

Ask question

All categories

3 years ago

12

At the carnival alana gave away 26 ride tickets to her friends. She now has 57 tickets left. Let x be the number of tickets that

alana started with.

1 answer:

iren2701 [21]3 years ago

5 0

x-26=57
+ 26 26
__________
x=83
Alana started with 83 tickets.
Is this what you mean?
I hope so! :)

You might be interested in

In how many ways can four aces be drawn from a deck of cards? (Order is not important.)

natka813 [3]

Let's imagine that we have in the deck 52 cards. So what we need to do is to make a kind of "progression".So let me explain this by doing the procedure. we can use is like this:
4/52 * 3/51 * 2/50 * 1/49 =
<span>24 / 6497400 = </span>
<span>1 / 270725 </span>
<span>3.694e-6</span>

8 0

2 years ago

PLEASE HELP! WILL GIVE BRAINLIEST!!!

lilavasa [31]

Answer:

C.1/4 the pqr is the prime of factor by negative comma 1

3 0

2 years ago

Read 2 more answers

Find the difference between (2x-5) and (3-9x)

BARSIC [14]

I think it might be 11x-8
I'm really sorry if that's wrong

4 0

3 years ago

In a translation, the line segments that connect each point on the preimage to the corresponding point on the image are not only

makvit [3.9K]

Parallel because congruency and parallelism is somehow similar and that word fits in perfectly

5 0

3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

2 years ago