Answer:
8
Step-by-step explanation:
x + 26 = 18
x = 18 - 26
x = 8
So basically you want to find the total percentage of cakes that weren't sponge cakes first.
You already have the 50% of party cakes. For the 1/5 percent of fruit cakes, you can multiply it by 20/20, to get 20/100, and then just take the 20 to get 20% that were fruit cakes.
Now you can just add the percentages together.
50% + 20% = 70%
So now you know 70% weren't sponge cakes, out of 100%.
So here you can just subtract 70% from 100% to figure out the remaining part of 100%, which must be sponge cakes.
100% - 70% = 30%
So 30% of the cakes were sponge cakes.
Answer:
a = 1 b = -1.
Step-by-step explanation:
Suppose the quotient when you divide x^4+x^3+ax+b by x^2 + 1 is
x^2 + ax + b then expanding we have:
(x^2 + 1)(x^2 + ax + b)
= x^4 + ax^3 + bx^2 + x^2 + ax + b
= x^4 + ax^3 + (b + 1)x^2 + ax + b Comparing this with the original expression:
x^4 + x^3 + 0 x^2 + ax + b Comparing coefficients:
a = 1 and b+ 1 = 0 so b = -1.
Answer: Is true sometimes.
Step-by-step explanation:
I guess that here we have two matrices, A and B, that are nxn.
We can see that if those matrices can conmutate, then we can try it with some simple matrices.
![A = \left[\begin{array}{ccc}1&0\\0&-1\end{array}\right] . B = \left[\begin{array}{ccc}2&0\\1&1\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D%20.%20B%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
Here, we would have that:
![AB = \left[\begin{array}{ccc}2&0\\-1&-1\end{array}\right]](https://tex.z-dn.net/?f=AB%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%5C%5C-1%26-1%5Cend%7Barray%7D%5Cright%5D)
![BA = \left[\begin{array}{ccc}2&0\\1&-1\end{array}\right]](https://tex.z-dn.net/?f=BA%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%5C%5C1%26-1%5Cend%7Barray%7D%5Cright%5D)
You can see that AB and BA are different, then the statement is not always true.
But it is true sometimes, if A or B are the identiti, then I*A = A*I, in this case would be true.
It is also true if A and B are diagonal matrices, let's prove it:
![A = \left[\begin{array}{ccc}a&0\\0&b\end{array}\right] , B = \left[\begin{array}{ccc}c&0\\0&d\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%260%5C%5C0%26b%5Cend%7Barray%7D%5Cright%5D%20%2C%20B%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc%260%5C%5C0%26d%5Cend%7Barray%7D%5Cright%5D)
![AB = \left[\begin{array}{ccc}ac&0\\0&bd\end{array}\right] = BA](https://tex.z-dn.net/?f=AB%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dac%260%5C%5C0%26bd%5Cend%7Barray%7D%5Cright%5D%20%3D%20BA)
Rational, 3.4 can be written as a fraction and is noncontinuous
"An irrational number <span>can be written as a decimal, but not as a fraction. An </span>irrational number<span> has endless non-repeating digits to the right of the decimal point."</span>
