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try graphing

Are you staking him?

4 times.

Let t minutes be the time needed.

Mike is traveling at \(\dfrac{71}{60}\text{ mi/min}\) and Ike is traveling at \(1\text{ mi/min}\).

1 minute after Ike passed the one-mile mark, Mike passes the one-mile mark. But in that one minute that Mike used to catch up to Ike, Ike has traveled an extra 1 mile. So at that specific moment:

|-------------------------|------------------------|

            1 mi            Mike        1 mi            Ike

t minutes after that moment, total distance travelled by Mike is \(\left(1 + \dfrac{71}{60}t\right)\) miles and total distance travelled by Ike is \((2 + t)\) miles.

When Mike catches up to Ike, the total distance traveled will become the same. so \(1 + \dfrac{71}{60}t = 2+ t\). Please solve for t on your own.

Then (t + 2) minutes is the answer.

Let x = 3^c. Then \(3x^2 = 28x\). You can use quadratic formula or factorization to solve for x. Can you take it from here?

Percentage increase = \(\dfrac{\text{new} - \text{old}}{\text{old}} \times 100\%\). Plug in new = 119 and old = 85 and you will get the answer.

https://web2.0calc.com/questions/i-need-help-fast_10#r1

Same question here

\((3x^3 - 3x + 4) - (x^3 - 2) = 3x^3 - 3x + 4 - x^3 + 2\)

Now you just collect the like terms and it's done. Remember to order your terms in descending order of power.

When \(x=0\), \(y = 0 \).

When \(x = 4\), \(y = 12 \)

The formula for the rate of change is \(\text{rise} \over \text{run}\), where rise is the change in y-coordinate, and run is the change in x-coordinate.

Can you take it from here? 

Oh course it is is schoolwork.  Almost every question here is schoolwork.

There are 52 cards in a standard deck.

probabillity of selecting a diamond:1/4

prob. landing on heads: 1/2

1/4 *1/2=1/8

(Pls text me if it's incorrect!)

Great work Catmg and Max :)

I will do (a) for you as an example.

x + floor(x) = 1

x = 1 - floor(x), which is an integer.

We conclude that x is an integer. Then floor(x) = x.

x + x = 1

2x = 1

x = 1/2 (rejected)

1/2 is rejected because it is not an integer.

There are no such x >= 0 such that f(x) = 1.

(b) and (c) are similar, and there are infinitely many possible values of f(x) for 0 <= x <= 10, if you consider all real numbers in that range.

10 * 11 * 12 * ... * 25 = (10 * 15 * 20 * 25) * (other numbers)

The other numbers don't have any factor of 5. We consider 10 * 15 * 20 * 25.

10 * 15 * 20 * 25 = 5^5 * 2^3 * 3

Pick two even numbers from (other numbers) to make 5^5 * 2^5 = 10^5.

There will be 5 zeros at the end.

Using binomial theorem, \(\newcommand{\dsum}{\displaystyle\sum} (2y - 7)^5 = \dsum_{k = 0}^5 \binom{5}k (2y)^k (-7)^{5 -k}\)

To find the term in y^4, the power needs to be 4. So let k = 4:

\(\text{term in }y^4 = \displaystyle\binom{5}4 (2y)^4 (-7)^1 = -560y^4\)

Coefficient of y^4 = -560.

Are you sure this isn't schoolwork...

We can rewrite the second equation as: \(5y=ax+11+x\)

We then factor this into: \(5y=x(a+1)+11\)

We now multiply the first equation by 5 to get: \(5y = 55x+ 14 \)

Recall that for a system of equations to have no solutions, the two lines must have the same slope. 

The slope is the co-efficient of x, so we have the equation: \(55 = 1+a\), where we can solve for a. 

Can you take it from here?

Rotate \(\triangle ADC\) so \(AD\) lies on \(DC\) and \(DC\) lies on \(DB\). 

Because of similar triangles, \(AD:DC = DC: DB = 4:6\). 

We now have this pair of ratios to solve for: \(4: 6 = 6: \text{__}\)

Can you solve it now?

The interior angles of a triangle sums up to 180 degrees, so \(C = 180^\circ - 42^\circ - 75^\circ = 63^\circ\).

Then you just use a calculator to calculate \(\sin C = \sin 63^\circ\).

Hints:

1) How can you relate the rate of change in area to the rate of change in radius?

2) What is the radius when the area is 10 mi^2? This is the value of the radius you need to substitute when you substitute area = 10 mi^2.

When \(f(x) \geq 4\), \((x^2 - 4 \geq 4 \text{ and }x \geq 4) \text{ or } (x^2 + 4 \geq 4 \text{ and }x < 4)\), which is \(x < 4\text{ or }x \geq 4\), which means f(x) is always greater than or equal to 4.

Then f(f(x)) = (f(x))^2 - 4 by definition of f.

Case 1: x >= 4

\((x^2 - 4)^2 - 4 = 5\\ x^4 - 8x^2 + 16 - 4 - 5 = 0 \\ x^4 - 8x^2 + 7 = 0\\ (x^2 - 7)(x^2 - 1) = 0\\ x = \pm \sqrt 7 \text{ (rej.) or }x = \pm 1 \text{ (rej.)}\)

The roots are rejected since none of them are greater than 4.

Case 2: x < 4

\((x^2 + 4)^2 - 4 = 5\\ x^4 + 8x^2 + 16 - 4 - 5 = 0\\ (x^2 + 1)(x^2 + 7) = 0\\ \text{No real solution}\)

Therefore, there are no value of x that would make f(f(x)) = 5.

Your answer looks very cool, but I wanted to try to see if it can be solved with calculus too because I haven't learned it yet. :))

m = maximum of ab(a-b)

x = a - b (like the hint said)

9 = a + b

a = (9+x)/2

b = (9-x)/2

m = maximum of (9+x)/2*(9-x)/2*x

Note: x has to be be nonegative since if b > a, and a and b are positive integers, ab(a-b) will be negative.

This is a cubic with roots at -9, 0, and 9.

By plugging in points and drawing a sketch, we can get an idea of what the cubic looks like. 

The graph approaches 0 from the positive side from as x approaches -9 from the negative side. 

Then, the graph crosses the x axis at -9, and goes down, but then goes back up to the x axis at 0. 

After that, the graph dips up but returns back to 0 at x = 9, and just continues going down.

So our maximum is at the turning point between 0 and 9, this is the distance m. 

Then we can use the method written here:

https://www.themathdoctors.org/max-and-min-of-a-cubic-without-calculus/

=^._.^=

Sketch of solution:

Suppose the rate at which water is being pumped into the tank is k cm^3 / min. Then the net rate of increase of volume is (k - 6700) cm^3 / min.

You can do what I did in the previous problem (considering the largest cross-section) to find the radius of the water surface when the height is h cm. Draw a diagram to help you visualize the problem. When you have that, you can express the volume V entirely in terms of h. Differentiate both sides with respect to time and then you can relate the rate of change in volume to the rate of change in water level. That rate of change in volume when h = 2 is exactly (k - 6700) cm^3 / min.

Please tell me if you encounter problems solving it according to my sketch.

This problem is very much like this one here. If you understood how to do that one, you know how to do this one.

https://web2.0calc.com/questions/domain_89573