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3 years ago

Find LJ and mGM if KP=PL and GH=36 ft

Mathematics

1 answer:

Aloiza [94]3 years ago

6 0

Answer: the answer to lg is 18

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Whats an approximate equation of a line that passes through the points (-4.3,2.3) and (1.3,-7.4)

Wewaii [24]

Use the 2 points to find the gradient of the line
Gradient = (y - y1)/(x - x1), y and y1 are the two different y values.
(2.3 - - 7.4)/(-4.3 - 1.3) = -97/56 = -1.732
Note: y and x both come from the same coordinate, and y1 and x1 also come from the same coordinates - (x , y), (x1 , y1)

Use the following to find the equation (x, x1, y, and y1 are not the same as the first part)
y - y1 = m(x - x1)

Where x2 and y2 is an intersection (one of the coordinates you used) and m is the gradient you found.
So...
y - 2.3 = -1.732(x - - 4.3)
You can simplify this if you are required to.

4 0

3 years ago

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

nordsb [41]

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$

5 0

3 years ago

You can do both problems if want

Agata [3.3K]

A= π r2 and v=3.14d

---Volume(v=3.14d):

3.14(15) equals 47.1

47.1 rounds to 47

nearest hundreth would be 0 that rounds to .1 or .10,which rounds to 47
v=47

---Area(π r2)

3.14(7.5)^2 is 3.14(56.25)
176.625

176.625 rounded to the nearest hundreth is 176.630.

Answer:

The area of a circle with the diameter of 15 is 176.630.

4 0

3 years ago

1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

8 0

2 years ago

The equation of a line that goes through the points (0,0) and (300000,365)

olga_2 [115]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 0 &,& 0~) 
% (c,d)
&&(~ 300000 &,& 365~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{365-0}{300000-0}\\\\\\ \cfrac{365}{300000}\implies \cfrac{73}{60000}$

$\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-0=\cfrac{73}{60000}(x-0)\implies y=\cfrac{73}{60000}x$

4 0

3 years ago