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3 years ago

How do u do parabola

Mathematics

1 answer:

alekssr [168]3 years ago

4 0

Answer:

Step-by-step explanation:

Finding the vertex or the end of the shape.

Find the y -intercept, (0,f(0)) ( 0 , f ( 0 ) ) .

Solve f(x)=0 f ( x ) = 0 to find the x coordinates of the x -intercepts if they exist.

Make sure that you've got at least one point to either side of the vertex. ...

Sketch the graph.

You might be interested in

What is the slope-intercept form of a line that contains the point (0,3) and a slope=-2

katen-ka-za [31]

Answer:

C. y = -2x + 3

Step-by-step explanation:

The equation of a line in slope-intercept form is given as y = mx + b.

m = slope, while b = y-intercept.

We already know that the slope, m, = -2.

We are given a point also as (0, 3). This means, at x = 0, y = 3.

Therefore, the y-intercept, b, = 3, because, the line would intercept the y-axis of the graph at y = 3, and at that point, x = 0.

Thus,

Slope (m) = -2,

y-intercept (b) = 3

Plug these values into y = mx + b

✅The equation of the line in slope-intercept form would be:

y = -2x + 3

7 0

3 years ago

1/8 divided by 3/4 in simplest form

Eddi Din [679]

(1/8) / (3/4) =
1/8 * 4/3 =
4/24 reduces to 1/6 <===

5 0

3 years ago

Read 2 more answers

2 2/3 x 3 3/8=? <br> Please help me

lara31 [8.8K]

Answer:

Step-by-step explanation:

2 2/3 = 8/3

3 3/8 = 27/8

8/3 * 27/8 = 216/24

= 9

3 0

3 years ago

Angle θ is in standard position. If (8, -15) is on the terminal ray of angle θ, find the values of the trigonometric functions.

notsponge [240]

ANSWER

$\sin( \theta) = - \frac{15}{17}$

$\csc( \theta) = - \frac{17}{15}$

$\cos( \theta) = \frac{8}{17}$

$\sec( \theta) = \frac{17}{8}$

$\tan( \theta) = - \frac{15}{8}$

$\cot( \theta) = - \frac{8}{15}$

EXPLANATION

From the Pythagoras Theorem, the hypotenuse can be found.

${h}^{2} = 1 {5}^{2} + {8}^{2}$

${h}^{2} = 289$

$h = \sqrt{289}$

$h = 17$

The sine ratio is negative in the fourth quadrant.

$\sin( \theta) = - \frac{opposite}{hypotenuse}$

$\sin( \theta) = - \frac{15}{17}$

The cosecant ratio is the reciprocal of the sine ratio.

$\csc( \theta) = - \frac{17}{15}$

The cosine ratio is positive in the fourth quadrant.

$\cos( \theta) = \frac{adjacent}{hypotenuse}$

$\cos( \theta) = \frac{8}{17}$

The secant ratio is the reciprocal of the cosine ratio.

$\sec( \theta) = \frac{17}{8}$

The tangent ratio is negative in the fourth quadrant.

$\tan( \theta) = - \frac{opposite}{adjacent}$

$\tan( \theta) = - \frac{15}{8}$

The reciprocal of the tangent ratio is the cotangent ratio

$\cot( \theta) = - \frac{8}{15}$

8 0

3 years ago

Read 2 more answers

Can two box plots have the same range and iqr and yet represent completely diff data.. explain

Mila [183]

Yes it is possible. Consider the following scenarios

Scenario A:
Min = 5
Q1 = 10
Median = 12
Q3 = 18
Max = 22

The IQR is equal to the difference of Q3 and Q1
IQR = Q3-Q1 = 18-10 = 8
The range is the difference of the min and max
Range = Max - Min = 22 - 5 = 17

So in summary for scenario A, we have
IQR = 8
Range = 17

----------------------------------------------------------

Now consider another scenario, call it scenario B, where

Min = 100
Q1 = 102
Median = 105
Q3 = 110
Max = 117

I claim that the IQR and Range for scenario B is going to be the same as in Scenario A. Let's find out

IQR = Q3 - Q1 = 110 - 102 = 8
Range = Max - Min = 117 - 100 = 17

So
IQR = 8
Range = 17
which is identical to scenario A.

-------------------------------------------------------------------------------

Scenario B has completely different data than scenario A, yet the IQR and Range are equal to scenario A's counterparts. This shows that it is possible to have 2 completely sets of data yet have the same IQR and range.

The wrap up here, and the answer to the question, is "yes it is possible" with the explanation given above.

4 0

3 years ago