A= 1/2bh
fill in formula with known numbers
42= 1/2(7)h
multiply numerators on right side
42= (1*7)/2h
42= 7/2h
divide both sides by 7/2
42 ÷ 7/2= h
to divide fractions, multiply by reciprocal/inverse of 7/2
42 * 2/7= h
multiply numerators
(42 * 2)/7= h
84/7= h
12 inches= h
ANSWER: The height is 12 inches.
Hope this helps! :)
A parabolic function's key characteristic is either having 2 x-intercepts or 2 y-intercepts. That is the reason why the standard form of parabolic functions are:
(x-h)^2 = +/- 4a(y-k) or (y-k)^2 = +/- 4a(x-h), where
(h,k) is the coordinates of the vertex
4a is the lactus rectum
a is the distance from the focus to the vertex
This is also called vertex form because the vertex (h,k) is grouped according to their variable.
Since we don't know any of those parameters, we'll just have to graph the data points given as shown in the picture. From this data alone, we can see that the parabola has two x-intercepts, x=-4 and x=-2. Since it has 2 roots, the parabola is a quadratic equation. Its equation should be
y = (x+4)(x+2)
Expanding the right side
y = x²+4x+2x+8
y = x²+6x+8
Rearrange the equation such that all x terms are on one side of the equation
x²+6x+___=y-8+___
The blank is designated for the missing terms to complete the square. Through completing the squares method, you can express the left side of the equation into (x-h)² form. This is done by taking the middle term, dividing it by two, and squaring it. So, (6/2)²=9. Therefore, you put 9 to the 2 blanks. The equation is unchanged because you add 9 to both sides of the equation.
The final equation is
x²+6x+9=y-8+9
(x+3)²=y+1
Answer:
d. - 15
Step-by-step explanation:
The equation for direct variation is
y = kx
We know y =3 when x = -9
3 = k(-9)
Divide by -9
3/-9 = -9k/-9
-1/3 = k
The equation is then
y = -1/3 x
We know y = 5
5 = -1/3 x
Multiply each side by -3
-3*5 = -3*-1/3x
-15 = x
Answer:
Step-by-step explanation:
Answer:
- 45/1024
- 1/4
- 15/128
- 193/512
- 9/512
Step-by-step explanation:
There are 2^10 = 1024 bit strings of length 10.
a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits
p(2 1-bits) = 45/1024
__
b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.
p(b0=0 & b9=0) = 1/4
__
c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.
p(7 1-bits) = 120/1024 = 15/128
__
d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits
p(more 0 bits) = 386/1024 = 193/512
__
e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.
p(2 1-bits | first is a 1-bit) = 9/512