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Zinaida [17]
3 years ago
12

Please help and explain!

Mathematics
1 answer:
Scrat [10]3 years ago
5 0
K is the slope of the line formed by connecting the points.
Use slope formula:
\frac{y_2 - y_1}{x_2 - x_1}
where the 2 points are the end points of the line (First and last going Left to right)
(x_1,y_1) = (2,7)  \\ (x_2,y_2) = (8,28)
Substitute into slope formula
k= \frac{28-7}{8-2} = \frac{21}{6} = \frac{7}{2}
You can check if this is correct by going back to graph and going "up 7" and "over 2" to get from one point to the next.
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Can someone please help me on this question I'm already having a bad day and I just want to get this stupid math over with
balu736 [363]

Answer:

Option A

Step-by-step explanation:

To check which function satisfies the graph, substitute the points. Points: $ (1, 12) $ is on the graph. Substitute it on the functions given in the options.

Option A: $ f(1) = 20 { (\  \frac{3}{5} )\ }^1 $

$ \implies 4 \times 3 = 12 $

Other options do not fit this $ (1,12) $. Hence Option A is our answer.

5 0
3 years ago
I think of a number multiply it by two subtract 5 and multiply the result by 3
Semmy [17]

Answer:

=)

Step-by-step explanation:

You could do 40 x 2 = 80 - 5 = 75 x 3 = 225

So if this is right, The answer would be 225

7 0
2 years ago
What is the exact value of cos (67.5°)?
babymother [125]

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill

cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}

8 0
2 years ago
In this particular arithmetic sequence, a6 = 37 and a17 = 114. What is the value of a23?
fgiga [73]
37=a+5d
114=a+16d
_________
114-37=16d-5d
77=11d
7=d
_________
37=a+5×7
37-35=a
2=a
________
a23=a+(23-1)×d
a23=2+(23-1)×7
a23=156
8 0
3 years ago
The directrix of a parabola is y = 4. Its focus is (2,6).
puteri [66]

Answer:

Equation in picturee

Step-by-step explanation:

8 0
3 years ago
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