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sammy [17]
3 years ago
8

HELP PLEASE NEED BEFORE TOMORROW MORNING

Mathematics
1 answer:
9966 [12]3 years ago
5 0
Okay YAS, I loove linear equations... moving on. 

OKAY, so I'm going to start with Q2 (not 100% sure how to work out 1, but lets use process of elimination). Firstly, the rule is y=mx+c So, start with the first line y=2x-3, from this the first thing you need to do is find the y-intercept (m) and the gradient (c). So, this is very simple 2 is m and -3 is c (if you look at the rule it matches up). 
m=2/1(this is rise/run, rise is how much you go up by, and run is how much you go left or right by (left if positive, right if negative))
c=-3
Now, go to your graph, and look on the y line and go to -3. Then go left, 1, and up 2. IS THERE A LINE CROSSING THROUGH THIS POINT? Yes? Good. 
Now for y=-1/2+1. Looking at the rule again (y=mx+c), 
m=-1/2 (rise/run)
c=1
NOW, once again go to your graph, and look on the y line and go to 1 (in this case the line actually passes through this point), now go right 2 places and up 1. IS THERE A LINE? Yes? Good!

NOOOW, for Q3, use the same steps and you will have the answer. 
y=1/4x-2
m=1/4 (rise/run)
c=-2
Look at graph C, start at -2 on the y line, go left 4, and up 1? Yes? Line crossing through this point? good!!
Now for the 2nd one, 
y= -2/5x+4
m=-2/3
c=4
Look at the same graph start at 4 on the y line, go right 5 and up 2! Got it? There's a line crossing through this point, yes? YAY

Now I have no idea how to do Q1, but by process of elimination it's pretty obvious. I hope this is right! Okay, bye!! :D
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A field is to be fertilized at a cost of $0.08 per square yard. the rectangular part of the field is 95 yd long and the diameter
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Assuming that this field is rectangular with 2 semi-circles (one on both ends of the field), we must first calculate the total area of the field. this is done by adding the area of the rectangular portion and the circular portion. This is done below:

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Rectangular length = 95 yards
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Area of rectangle = length * width = 95 * 49
Area of rectangle = 4655 yd^2

The area of the 2 semi-circles can be obtained by treating both as one circle.
Area of circle = pi * (d/2)^2 = 3.1416 * (49/2)^2
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Question 15 (5 points)<br> Find the angle between u = &lt;7, -2&gt; and v= (-1,2&gt;.
Tema [17]

Answer:

Approximately 2.3127 radians, which is approximately 132.51^{\circ}.

Step-by-step explanation:

Dot product between the two vectors:

\begin{aligned}& u\cdot v \\ =\; & 7 \times (-1) + (-2) \times 2 \\ =\; & (-11) \end{aligned}.

Magnitude of the two vectors:

\begin{aligned} \| u \| &= \sqrt{{7}^{2} + {(-2)}^{2}} \\ &= \sqrt{53} \end{aligned}.

\begin{aligned} \| v \| &= \sqrt{{(-1)}^{2} + {2}^{2}} \\ &= \sqrt{5} \end{aligned}.

Let \theta denote the angle between these two vectors. By the property of dot products:

\begin{aligned} \cos(\theta) &= \frac{u \cdot v}{\|u\| \, \| v \|} \\ &= \frac{(-11)}{(\sqrt{53})\, (\sqrt{5})} \\ &= \frac{(-11)}{\sqrt{265}}\end{aligned}.

Apply the inverse cosine function {\rm arccos} to find the value of this angle:

\begin{aligned} \theta &= \arccos\left(\frac{u \cdot v}{\| u \| \, \| v \|}\right) \\ &= \arccos\left(\frac{(-11)}{\sqrt{265}}\right) \\ & \approx \text{$2.3127$ radians} \\ &= 2.3127 \times \frac{180^{\circ}}{\pi} \\ &\approx 132.51^{\circ}\end{aligned}.

8 0
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