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Ahat [919]
2 years ago
13

Graph y > -1/3x+5

Formula1" title="y > - \frac{1}{ 3 } x + 5" alt="y > - \frac{1}{ 3 } x + 5" align="absmiddle" class="latex-formula">
​

Mathematics
1 answer:
ad-work [718]2 years ago
6 0

Explanation:

The function is y>-\frac{1}{3} x+5

To graph the function, let us find the x and y intercepts.

To find x-intercept, let us substitute y=0 in the function y>-\frac{1}{3} x+5

\begin{aligned}0 &=-\frac{1}{3} x+5 \\-5 &=-\frac{1}{3} x \\x &=15\end{aligned}

Thus, the x-intercept is (15,0)

To find the y-intercept, let us substitute x=0, we get,

\begin{aligned}&y=-\frac{1}{3}(0)+5\\&y=5\end{aligned}

Thus, the y-intercept is (0,5)

The graph has no asymptotes.

To plot the points in the graph, we need to substitute the values for x in the function y>-\frac{1}{3} x+5, to find the y-values.

The points are (-2,5.667),(-1,5.333),(1,4.667),(2,4.333),(3,4). The image of the graph and table is attached below:

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These 2 questions confuse me.<br> Can anyone help with some 3D trigonometry?
Burka [1]

Answer:

Q2) 29 degree       as unrounded to nearest degree is 28.95 degree

Q3) 69 degree       as unrounded to nearest degree is 68.56 degree

Step-by-step explanation:

QU 2)

When they speak of plane we see ABCD and also see ABC

So we need the length of AB and BC to find the diagonal CA

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Tan x = tan-1 10.4/18.8 = 28.95099521 degree

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so we know one is much smaller than the other

We also know ACG angle is 90 degree and that angle from ABCD that meets line AG is the smaller angle.

Answer therefore must be 28.95 degree = or 29 degree

 

QU 3)

we are basically looking for angle where VB meets BC line or AVB meets ABC we have the slant length, so step 1 is find the height by first dividing square base by 2 then finding the height.

= 7.6/2 = 3.8 cm

Then Pythagoras

BV^2  -  1/2 BC =  height

10.4^2  - 3.8^2 = height

Height = sq rt 93.72 =9.68090905 = 9.7cm

Which means  V to midpoint VC = V to midpoint  AB

They are the same and the midpoints are 90 degree angles.

To find the required angle for VB + BCmidpoint or we wont be able to determine the right angle hypotenuse.

We do the same as last question determine the hypotenuse and where the angle sought is is where we use the trig function = adj/hyp

Because if it was the midpoint angle then it would be opp/adj like the question 1  so this time its cos of x.

cos x = adj/hyp = cos-1 (3.8/ 10.4) = 68.5687455

Answer is 68.56 degree

The reason we show the height is so we can check by doing opp/hyp

= sin of x = sin-1 (9.68090905/3.8) = 23.11171135

and 90 -23.11171135 = 66.8882887

= 67 degree

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= sin-1 (3.8/9.7) = 23 degree  90-23 = 67 degree

but when rounded to 10.4cm for slant we get the same

= sin-1 (3.8/10.4)

So we realise here trig functions -1  doesn't work on the same 90 degree angle for both lines that meet such 90 degree angle.

We try the sin-1 (10.4/ 9.68090905) = 68.5687455 = 69 degree

and that where the lines join away from the 90 degree angle we can always find true answer, and see it is a match with the first cos trig function we did.

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and smallest side acts as denominator for both trig functions.

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