Thank you so much CPhill
We are putting "g" into " f"
So
(x - 3)^2 + 1 =
x^2 - 6x + 9 + 1 =
x^2 - 6x + 10
I'm not sure whether there's one line of symmetry or 5...
See here : https://web2.0calc.com/questions/i-ve-been-stuck-on-this-problem-for-ages-is-it-actually-easy
Five ways with the face up five more ways with the face reflected to the back .... 10 ways
I think it might be, in which case I think there might be 50 solutions
Can anybody help with this problem?
Is it legal to flip the pentagon over, so that the back side of the cardboard is now facing up?
Double repost. I answered 2 and 3 here:
https://web2.0calc.com/questions/show-that-when-the-square-of-an-plsss-helpp#r1
A pentagon can be rotated 5 ways and still have it look the same. I hope this helps. I myself still am not sure of the solution yet.
This sum is 1 + 2 + 4 + 8 + 16 = 31.
31 is prime itself, so the answer is 31.
Add values of c such that when the graph is moved up, it has 4 intersections with the x-axis.
This question seems to be a repost. I answered this already.
https://web2.0calc.com/questions/equation-math-question-pls-help-gt-gt-gt-anyone
We are solving:
\(x(1.5) = 25 \quad \rightarrow x = \frac{25}{1.5} = \boxed{\frac{50}{3}}\)
\(x(2.5) = 42 \quad \rightarrow \quad x=\frac{42}{2.5} = \boxed{16.8}\)
\(87+213123+898 \cdot 1231231238- \frac{51235213}{5424554}\)
\((213210) + (1105645651724) - (9.445) = \boxed{1,105,645,864,924.555} \)
Let S be the weight of a small box and L be the weight of a large box. We can write a system of equations. Hope this helps.
We know the vertex, so we can write the equation in vertex form: \(y = a(x+1)^2 + 7\)
The y-intercept is (0, 10), so plug that point in: \(10 = a + 7\)
So, a is equal to 3, and we now have the equation: \(y = 3(x+1)^2 + 7\)
I. The range of both distributions is 7 - 0 = 7 hours, so this is true.
II. This is also true, since the seniors graph is skewed.
III. The median is 1 and the mean is roughly 1.667, so this is false.
1. \(10001^2 -9999^2 = (10001 + 9999)(10001 - 9999) = (20000)(2) = \boxed{40000}\)
2. \(\frac{-5+2i}{1+7i} \cdot \frac{1-7i}{1-7i} = \frac{-5+2i+35i-14i^2}{50} = \boxed{\frac{9+37i}{50}}\)
ABCD is a parallelogram, so angle x = w and angle y = z.
w = 180 - 53 = 127 degrees
x = w = 127 degrees
y = z = 53 degrees
Make sure you understand why this is true.
Start by evaluating O(3) in the middle and working outwards. It's a tedious problem, but straightforward.
O(3) = 3^2 = 9
N(9) = 2/sqrt(9) = 2/3
O(2/3) = 4/9
...
and so on.
BO = OD
7x + 4 = 18
7x = 18 - 4
7x = 14
x = 14 / 7 = 2
1. \((x+2)(x^2+3) = \boxed{x^3 + 2x^2 + 3x + 6}\)
2. \(x^2 - 9x - 36 = \boxed{(x-12)(x+3)}\)
3. \(3x^2 +14x - 5 = 0\)
(quadratic formula)
\(x = {-14 \pm \sqrt{196-4(3)(-5)} \over 2(3)} = {-14 \pm 16 \over 6}\)
\(\boxed{x=-5, -\frac{1}{3}}\)
If the degree of f + g = 1
Then we can le g be (at the least) something like 7x^4 - 3x^3
So g must be degree 4
Not too much here,wiseowl......
20 ( 1.25) (.40) = $10
??
\(\displaystyle x^{2}=(12-a)^{2}+b^{2}.\)
