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brilliants [131]

2 years ago

8

Kaleb went to a theme park with $25 to spend. He spent $5.25 on food and paid $4.00 for each ride. What was the greatest number

of rides Kaleb could have rode

1 answer:

Shtirlitz [24]2 years ago

8 0

Answer is 4 rides
19.75 after food so divide it by 4 to get your answer

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A village was founded four hundred years ago by a group of 20 people. In this village, the population triples every one hundred

Paha777 [63]

Answer:

Step-by-step explanation:

Treat this like compound interest: Use A = P(1 + r)^t.

Here, P is the initial population and A is 3 times that, or 3P. Since P = 20 people, 3P = 60 people,

and this population is reached after 100 years.

We need to determine r, substitute its value into the formula A = P(1 + r)^t, and then determine the population of the village after 400 years.

60 = 20(1 + r)^100

Simplifying, 3 = (1 + r)^100.

Taking the natural log of both sides,

ln 3 = 100 ln (1 + r), or

ln 3

ln (1 + r) = ---------------

100

= 1.0986 / 100 = 0.01986

We must solve this for r. Raising e to the power ln (1 + r), on the left side of an equation, and raising e to the power 0. 01986 on the right side, we get:

1 + r = 3, so r must = 2.

Now find the pop of the village today. Use the same equation: A = P (1+r)^t.

A = 20(1 +2)^4 (hundreds),

or

A = 20(3)^4, or

A = 81

The population after 400 years is 81.

8 0

2 years ago

Consider the conjecture that the sum of a rational number and an irrational number is ALWAYS irrational. To try to prove this co

miss Akunina [59]

Answer:

B) irrational. Since an irrational number cannot equal a rational number.

The premise that c is rational is false

Step-by-step explanation:

The assumption that c is rational means that the difference c-a must be rational. That difference is b, so b=c-a would mean that b must be rational. But b is defined to be irrational. Since an irrational number cannot equal a rational number, a contradiction arises and the assumption that c is rational must be false. Therefore the sum (c) must be irrational. (B)

6 0

2 years ago

Read 2 more answers

Here are the ingredients to make 12 cupcakes.

scoundrel [369]

Answer:

Mark needs 2 kg butter, 2 kg caster sugar, 14 eggs, and 1.5 kg flour.

Step-by-step explanation:

Alright, first you double the number of kids, as Mark wants each kid to have two cupcakes, not one. (Then add 80+80 to get 160)

Now add the adults to children: 160+152=312

now divide by 12 to see how many batches of cupcakes he needs to make.

312÷12=26 So he needs to make 26 batches

Now here are the ingredients needed to make 12 batches of cupcakes (I multiplied the amount of each ingredient by 26)

5200 g (5.2 kg) butter, 5200 g (5.2 kg) caster sugar, 104 eggs, and 6500 g (6.5 kg) flour.

now make the measurements the same as Mark's ingredients (or the other way around)

(to save time I am not showing each one, just the normal ingredients turned into kg and compared)

5 kg butter < 5.2 kg butter

5 kg caster sugar < 5.2 kg butter

90 eggs < 104 eggs

5 kg flour < 6.5 kg flour

Now do a little subtraction.....

5.2 kg-5 kg = 2 kg butter is needed....

5.2 kg - 5 kg = 2 kg caster sugar is needed....

104 - 90 = 14 eggs needed....

6.5 kg - 5 kg = 1.5 kg flour needed....

And there you go!!! Hope this helped!!! :D

4 0

2 years ago

logan has 20 action figures.he is shipping them to a friend.He can only fit 3 action figure in a box.How many boxes will he need

adoni [48]

20/3= 6 2/3
6.66 rounds to 7
7 boxes

6 0

2 years ago

Read 2 more answers

A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon

posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

$V(t)$ is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

$V(t)=100 + 2t$

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

The amount of salt in the tank when it is full of brine is

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

6 0

2 years ago