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2 years ago

7

Please answer these I’m desperate. It’s do tomorrow and I only have those left so please. Btw they are two separate questions. I

already answered have of it I just need to find the final answer.
Instructions: preform the operations to climb the pyramid multiply to move up the pyramid and divide to move down the pyramid.

1 answer:

luda_lava [24]2 years ago

5 0

Answer:

it's 3

Step-by-step explanation:

I did this stuff in 6th grade it was really easy

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Glenda and Fiero are at a restaurant eating meal that costs $30. They have a discount coupon that gets them 10% off the cost. If

yaroslaw [1]

The cost of meal = $$30$

The tip to the waiter = $8%$ %

Discount = $10%$ %

Cost = $30 + 0.08 * 30 - 0.1 * 30$

Cost = $30(1 + 0.08 - 0.1)$

Cost = $ $$29.4$ 0

8 0

1 year ago

A square has an area of 100 square meters.<br> What is the perimeter of the square?

LiRa [457]

Answer:

40 square meters

Step-by-step explanation:

Perimeter means to add up all sides.

area means one side times one side

10 times 10 makes 100 so i am guessing each side is 10

10 + 10 + 10 + 10 = 40

5 0

2 years ago

Suppose a football is kicked with an initial velocity of 82 ft/sec., at an angle of

satela [25.4K]

Answer:

The position P is:

$P = 87\^x + 75\^y$ ft <u><em> Remember that the position is a vector. Observe the attached image</em></u>

Step-by-step explanation:

The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity $s_0$ is:

$y(t) = y_0 + s_0t -16t ^ 2$

Where $y_0$ is the initial height = 0 for this case

We know that the initial velocity is:

82 ft/sec at an angle of 58 ° with respect to the ground.

So:

$s_0 = 82sin(58\°)$ ft/sec

$s_0 = 69.54$ ft/sec

Thus

$y(t) = 69.54t -16t ^ 2$

The height after 2 sec is:

$y(2) = 69.54 (2) -16 (2) ^ 2$

$y(2) = 75\ ft$

Then the equation that describes the horizontal position of the ball is

$X(t) = X_0 + s_0t$

Where

$X_ 0 = 0$ for this case

$s_0 = 82cos(58\°)$ ft / sec

$s_0 = 43.45$ ft/sec

So

$X(t) = 43.45t$

After 2 seconds the horizontal distance reached by the ball is:

$X (2) = 43.45(2)\\\\X (2) = 87\ ft$

Finally the vector position P is:

$P = 87\^x + 75\^y$ ft

6 0

2 years ago

George studied 3/4 hour on monday and 7/8 hour on tuesday. how long did he study altogether on both days?

pentagon [3]

$\dfrac{3}{4}+\dfrac{7}{8}=\dfrac{3\cdot2}{4\cdot2}+\dfrac{7}{8}=\dfrac{6}{8}+\dfrac{7}{8}=\dfrac{6+7}{8}=\dfrac{13}{8}=\dfrac{8+5}{8}=\dfrac{8}{8}+\dfrac{5}{8}=1\dfrac{5}{8}\\\\Answer:\boxed{1\frac{5}{8}\ hour.}$

5 0

3 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

2 years ago