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katrin2010 [14]
3 years ago
10

Tto mesoort

Mathematics
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

4000Pa

Step-by-step explanation:

Pressure is force(thrust) per unit area.

Pressure=Force/Area

P=F/A

A.T.Q

Applied force=800N

Area=0.2m²

Pressure=800/0.2=4000Pa

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SOMEONE PLEASE HELP MEEEEE3​
Darina [25.2K]
The answer is (3,4)
3 0
3 years ago
To the person who makes a graph and doesn't copy from another question
allochka39001 [22]

Answer:

H=9;8)

B=(5;4) (ball)

R=(7;0) (hit point)

B'=symetric of B axis perpendicular of x in R

B'=(7+(7-5);4)=(9;4)

Equation BR: y-4=(0-4)/(7-5)(x-5)==>y=-2x+14

Equation RB': y-4=(4-0)/(9-7)(x-9)==>y=2x-14

Is H a point of RB'? y=2x-14 : 8=? 2*9-14==>8=?4 No!

 

you will not make your putt

Step-by-step explanation:

7 0
3 years ago
Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
IrinaK [193]
We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


8 0
3 years ago
stephanie holds 30 cubes in a box. the cubes are 1/2 in each side. the box holds 2 layers with 15 cubes in each layer. what is t
TEA [102]

Answer:

3.75 Cubic Inches

Step-by-step explanation:

The volume of a single cube is 0.125 because the volume of a cube is its side length cubed, or to the third power. There are 30 cubes in the box, and 0.125 times 30 is 3.75

8 0
2 years ago
I really need help on this question.!
IrinaVladis [17]

Answer:

b = 135 \\ a + c = 45(ie \: 180 - 135)  \\ a =  \frac{45}{2}  \\ a = 22.5

4 0
2 years ago
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