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Feliz [49]
3 years ago
6

Write a polynomial function in standard form with zeros at 0, 1, and 2.

Mathematics
1 answer:
Nikitich [7]3 years ago
6 0

Answer:

the first one is correct

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Help me iv been trying from this morning
GalinKa [24]

Answer:

i did the first

Step-by-step explanation:

1st way

Standard form: a(X-h)²+k = ( -2/3X² -16/3X -32/3) +32/3 -17/3 = -2/3(X +4)² +5

y = -2/3*X^2-16/3*X-17/3

X = -4 ±√( 15/2) = -6.7386, or -1.2614

Axis of symmetry: X= -4; Vertex (maximum)=(h,k)=( -4, 5); y-intercept is (0,-5.66666666667)

two real roots: X=-1.2613872124776866 and -6.738612787477313

5 0
2 years ago
Read 2 more answers
Which box-and-whisker plot represents the data set?<br><br> 11, 6, 19, 14, 21, 7, 13, 15, 15
Katena32 [7]
The box and whisker plot is attached.

We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21

The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median).  This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median).  This is between 15 and 19:  (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.

We draw the middle line of the box at 14, the median.  We draw the left side of the box at the lower quartile, 9.  We draw the right side of the box at the upper quartile, 17.  From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.

4 0
3 years ago
A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

y-0.721=(-1)(x-0.235)\\\\y=-x+0.956

The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

5 0
2 years ago
PLEASE HELP AND EXPLAIN! Find the height of the triangle.
dlinn [17]
Hello!


So you need to find the sin of the angle:

sin20 = x / 10

10 x sin20 = x

x = 3.4

Hope this helps :))
7 0
3 years ago
Read 2 more answers
The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
maxonik [38]

Given:

The red figure dilated with a scale factor of s=\dfrac{1}{3} and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

Solution:

If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

In given problem, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

4 0
3 years ago
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