The product is equal to $4^{1/3} = \sqrt[3]{4}$, so a + b = 7.
The graph of the equation y = ax^2 + bx + c, where a, b, and c are constants, is a parabola with axis of symmetry x = -10. Find b/a.
According to The Vieta's Formula, we can know
that -b/2a = -10, and b = 20a, and b/a = 20
BD = √7
[ABCD] = 2√3
CE = [ABCD] / √7
you forgot the possibility that x^2-4x+2=+-1
In the diagram, triangles BDF and FCE have the same area. If DB = 2, BA = 4, and AE = 3, find the length of EC.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If [BDF] = [FCE] then [ABC] = [ADE]
[ADE] = 1/2(6 * 3) = 9
1/2[4 * (3 + EC)] = 9
EC = 1.5
triangle BDF and ACE have the same area
Except ACE is a straight line. Typo?
.
x =3 is one of the zeroes....this would be one of the legs of the triangle....the other would occur on the y axis and needs to be = |1| for the area (1/2 b*h) = 1.5
the y axis crossing occurs when x = 0
-1 = 5 (x-3)(x+k) sub in x = 0
-1/5 = (-3)(k)
k = 1/15
when y = +1 1 = 5 (0-3((0+k )
1/5 = - 3k
k = - 1/15
Here is a desmos graph:
Note that anything raised to the '0' power = 1
5^0 = 1 123^0 = 1
so the exponent = 0
x^2 -5x + 2 = 0
Quadratic Formula shows x = 1/2 ( 5 +-sqrt(17) ) sum = 5
Where did you get 90 for BAC?
On this forum, there is a calculator in the "Home" part, where you can input this question.
Make sure that you type the question in correctly. "^*" does not exist.
=^._.^=
Your question is not complete.
x^2+2y-9=-y^2+28x+19 re-arrange into circle form (x-h)^2 + (y-k)^2 = r^2
x^2 -28x + y^2 + 2y = 28 complete the square forx and y
(x-14)^2 + (y+1)^2 = 196 + 28 + 1
center = 14, -1 r = sqrt (196+28+1) I think you can finish !
if b = - 1/2 the x^14 terms will 'cancel out' leaving degree 10
No matter WHAT constant 'a' equals, the degree will never be greater than 14
a/b = 3.
Calculate sqrt(60x)*sqrt(12x)*sqrt(63x)*sqrt(42x). Express your answer in simplest radical form in terms of x.
Hello Guest!
\(\sqrt{60x}\cdot \sqrt{12x}\cdot \sqrt{63x}\cdot \sqrt{42x}\\ =\sqrt{60\cdot 12\cdot 63\cdot 42\cdot x^4}\\ =\sqrt{2^2\cdot 3\cdot 5\cdot 2^2\cdot 3\cdot 3^2\cdot 7\cdot 2\cdot 3\cdot 7\cdot x^4}\\ =\sqrt{2^5\cdot 3^5\cdot 5\cdot 7^2\cdot x^4}\\ =2^2\cdot 3^2\cdot 7\cdot x^2\cdot \sqrt{2\cdot 3\cdot 5}\)
\(\sqrt{60x}\cdot \sqrt{12x}\cdot \sqrt{63x}\cdot \sqrt{42x}=\color{blue}252x^2\cdot\sqrt{30}\)
!
Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.
You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3. Using long division, you get the equations
-10a + 6b - 3c = 0
4a - 3b + 2c = 0
-a + b - c + 2 = 0
==> a = 6, b = 16, c = 12
So f(x) = 6x^5 + 16x^4 + 12x^3.
Thank you!
I entered that answer and it said it's wrong.
Angle EDC = arctan (sqrt(3)/2) = 40.9 degrees
then angle ECD = 180 - 90 - 40.9 = 49.1 degrees
cos (ECD) = adj / hyp
cos (49.1) = EC / 2
EC = 1.3 units
First ....add the two of the equations together......the 't' will now be gone and you can solve for 's'
use this value of 's' in either of the equations to solve for 't'
Assume that the radius of the Earth is r feet.
She travels the circumfrence of the Earth, which is 2*pi*r.
However, when we count her head instead of her feet, the radius gets increased by 5 feet.
(2*pi*(r+5)) - (2*pi*r) = 10pi
The polynomial can be factored:
(2x - 3i)(ix - 1) = 0
Setting each factor to zero and solving yields the two roots:
2x - 3i = 0
2x = 3i
x = 3i/2
ix - 1 = 0
ix = 1
x = -i (where we use the fact that 1/i = -i)
!
