Hey if someone could please help me with this i waould be very greatful

Kipish [7]

1. h = -16t² + vt + c
⇒ vt = 16t² + h - c
⇒ v = 16t + (h-c)/t

2. c = 5 ft
t = 3s
h = 131 ft

v = 16*3 + (131-5)/3 = 48 + 42 = 90 ft/s

when t=0, velocity is not defined. That means at t=0, there is no velocity of the t-shirt or it is 0.

5 0

2 years ago

Read 2 more answers

What is the equation of the line that passes through the points (-4,1) and (4-1)

Nitella [24]

Answer:

y=-1/4x

Step-by-step explanation:

METHOD 1: The basic slope-intercept equation is y=ax+b. To achieve the slope with two points, this is the equation:

$m = \frac{y2-y1}{x2-x1}$

All you have to do is insert points (-4, 1) and (4, -1) into their corresponding locations.

$m = \frac{(-1)-1}{4-(-4)}$

With this equation, you will be able to find the slope. Then you can simply graph it to connect the two dots.

METHOD 2: Simply connect the two dots with a straight line (use a ruler or something with a straight edge.) Then count how many spaces up/down and left/right. If the line goes from top to bottom from left to right, it is a negative slope (hence, the -1/4.)

7 0

2 years ago

determine the point of intersection of the following pair of lines ............2x-3y-4=-13 and 5x=-2y+25. 3x+2y-7=0 and 2x=12-5y

grandymaker [24]

Answer:

(x, y) = (3, 5)
(x, y) = (1, 2)

Step-by-step explanation:

A nice graphing calculator app makes these trivially simple. (See the first two attachments.) It is available for phones, tablets, and as a web page.

__

The usual methods of solving a system of equations involve elimination or substitution.

There is another method that is relatively easy to use. It is a variation of "Cramer's Rule" and is fully equivalent to elimination. It makes use of a formula applied to the equation coefficients. The pattern of coefficients in the formula, and the formula itself are shown in the third attachment. I like this when the coefficient numbers are "too messy" for elimination or substitution to be used easily. It makes use of the equations in standard form.

_____

1. In standard form, your equations are ...

2x -3y = -9
5x +2y = 25

Then the solution is ...

$x=\dfrac{-3(25)-(2)(-9)}{-3(5)-(2)(2)}=\dfrac{-57}{-19}=3\\\\y=\dfrac{-9(5)-(25)(2)}{-19}=\dfrac{-95}{-19}=5\\\\(x,y)=(3,5)$

__

2. In standard form, your equations are ...

3x +2y = 7
2x +5y = 12

Then the solution is ...

$x=\dfrac{2(12)-5(7)}{2(2)-5(3)}=\dfrac{-11}{-11}=1\\\\y=\dfrac{7(2)-12(3)}{-11}=\dfrac{-22}{-11}=2\\\\(x,y)=(1,2)$

_____

Note on Cramer's Rule

The equation you will see for Cramer's Rule applied to a system of 2 equations in 2 unknowns will have the terms in numerator and denominator swapped: ec-bf, for example, instead of bf-ec. This effectively multiplies both numerator and denominator by -1, so has no effect on the result.

The reason for writing the formula in the fashion shown here is that it makes the pattern of multiplications and subtractions easier to remember. Often, you can do the math in your head. This is the method taught by "Vedic maths" and/or "Singapore math." Those teaching methods tend to place more emphasis on mental arithmetic than we do in the US.

7 0

3 years ago

Solve the system of equations.

Helga [31]

The answer is x=1 y=-1 and z=1. If you plug in the variables you get
4-3+5 and that equals 6. So that’s the answer

7 0

2 years ago

Look at the figure XYZ in the coordinate plane. Find the perimeter of the figure rounded to the nearest tenth.

ki77a [65]

the distance form X to Y is clearly -6 to 0 is 6 units, and 0 to 8 is 8 units, so 6 + 8 = 14 units.

now, for XZ and ZY we can simply use as stated, the distance formula to get those and then add them all to get the perimeter.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ X(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ XZ=\sqrt{[5-(-6)]^2+[8-2]^2}\implies XZ=\sqrt{(5+6)^2+(8-2)^2} \\\\\\ XZ=\sqrt{121+36}\implies \boxed{XZ=\sqrt{157}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad Y(\stackrel{x_2}{8}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ZY=\sqrt{(8-5)^2+(2-8)^2}\implies ZY=\sqrt{9+36}\implies \boxed{ZY=\sqrt{45}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{perimeter}{14+\sqrt{157}+\sqrt{45}}\qquad \approx \qquad 33.2$

6 0

2 years ago