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Natali [406]
3 years ago
15

Determine an equation of a quadratic function with x intercepts of -3 and 5 that passes through the point A(4,-21)

Mathematics
1 answer:
erastova [34]3 years ago
4 0

Answer: y = 21x + -105 or y = 21x - 105


Step-by-step explanation:

Our equation is going to be in the format of y = mx+b, where m is the slope, and b is the y-intercept.

For this, you can find the slope by doing

(y₁-y₂)÷(x₁-x₂)

So we just sub in one of the x intercept points and the point that we already have, so let's do (-21-0)/(4-5). That is equal to 21, therefore, 21 is our slope.

So we have y = 21x + b, so now we can sub in one of our points for y and x in the equation. Let's do (4,-21).

-21 = 21(4) + b

-21 = 84 + b

-105 = b

So our equation is y = 21x + -105, meaning we have a slope of 21 and a y intercept of -105.


Edit: you can also do y = 21x - 105

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Which sentence describes why polygon MNOP is congruent to polygon JKLP?
svetlana [45]
The segment MN is equal in size to the segment JK, the segment NO is equal in size to the segment KL, the segment OP is equal in size to the segment LP.
7 0
3 years ago
If log9 x = 3/2 what is the value of x ? 1 3/2 2 3 27/2 4 . 27
sweet-ann [11.9K]

log 9 X= 3/2

x= 3.5

or x= 7/2

5 0
3 years ago
Point A is located at (−214,14). Point B is located at (234,14). What is the distance between point A and point B?
Sunny_sXe [5.5K]

Answer:

10, 14

Step-by-step explanation:

take the x values and plug into mid point formula and then divide by 2

so... (-214 + 234) divided by 2 = 10

do the same for the y values

so ... (14 + 14) divided by 2 = 14

so ... 10,14

7 0
2 years ago
Read 2 more answers
Question in picture<br><br>A) 16/63<br><br>B)-16/63<br><br>C) 63/16<br><br>D) -63/16
Orlov [11]
ANSWER
\tan(x + y) =  -  \frac{63}{16}


EXPLANATION


We were given that,

\csc(x)  =  \frac{5}{3}

This implies that,

\sin(x)  =  \frac{3}{5}

We use the Pythagorean identity

\sin^{2} (x)  +  \cos^{2} (x)= 1
to get,


\cos(x)  =  \sqrt{1 - ( { \frac{3}{5} })^{2}}  =  \frac{4}{5}


We were also given that,


\cos(y)  =  \frac{5}{13}

This means that,


\sin(y)  =  \sqrt{1 -  {( \frac{5}{13}) }^{2} }  =  \frac{12}{13}

This is because,


0 <  \: x \:  <  \frac{\pi}{2}


0 <  \: y \:  <  \frac{\pi}{2}

This angles are in the first quadrant so we pick the positive values.

\tan(x + y)  =  \frac{ \sin(x + y) }{ \cos(x + y) }


\tan(x + y)  =  \frac{ \sin(x ) \cos(y)   +  \sin(y)  \cos(x) }{ \cos(x) \cos(y)  -  \sin(x)  \sin(y) }



\tan(x + y)  =  \frac{  \frac{3}{5}   \times  \frac{5}{13}  +   \frac{12}{13}   \times  \frac{4}{5}  }{  \frac{4}{5}  \times  \frac{5}{13}   -   \frac{3}{5}  \times  \frac{12}{13}  }



\tan(x + y) =  -  \frac{63}{16}

The correct answer is D
4 0
3 years ago
How do you solve this
Dmitriy789 [7]

By definition of the trig functions,

\sin30^\circ=\dfrac y{16}

\cos30^\circ=\dfrac x{16}

We have \sin30^\circ=\dfrac12, so y=8, and \cos30^\circ=\dfrac{\sqrt3}2, so x=8\sqrt3.

Just to check: we should have

x^2+y^2=16^2

by the Pythagorean theorem. Indeed,

8^2+(8\sqrt3)^2=4\cdot8^2=2^2(2^3)^2=2^8=(2^4)^2=16^2

6 0
3 years ago
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