Answer:
6
Step-by-step explanation:
288 = 300-2x
Answer:
No such pair of integers. Hence, no solution.
Step-by-step explanation:
As the only factors of
58 are {
1
,
2
,
29
,
58
}
, there are no two consecutive integers whose product is 58
.
Answer:
The value of c that will result in a perfect square trinomial is (3)^2 or 9
The perfect square trinomial is: ![(w+3)^2=139](https://tex.z-dn.net/?f=%28w%2B3%29%5E2%3D139)
Step-by-step explanation:
We need to determine the value of c that will result in a perfect square trinomial.
![w^2+6w+ c =130+c](https://tex.z-dn.net/?f=w%5E2%2B6w%2B%20c%20%3D130%2Bc)
Perfect square trinomial are of form: ![a^2+2ab+b^2=(a+b)^2](https://tex.z-dn.net/?f=a%5E2%2B2ab%2Bb%5E2%3D%28a%2Bb%29%5E2)
Now, the equation given is:
![w^2+6w+ c =130+c](https://tex.z-dn.net/?f=w%5E2%2B6w%2B%20c%20%3D130%2Bc)
Looking at the term 6w, we can write it as 2(w)(3)
We are given: a = w, 2ab = 2(w)(3) so, b will be: (3)^2
So, we will be adding (3)^2 on both sides
![w^2+6w+ c=130+c\\w^2+2(w)(3)+ (3)^2 =130+(3)^2\\The\:left\:side\:becomes: a^2+2ab+b^2\\We\:can\:write\:it\:as: (a+b)^2\\We\:have\:a=w\: and\: b=3\\(w+3)^2=130+9\\(w+3)^2=139](https://tex.z-dn.net/?f=w%5E2%2B6w%2B%20c%3D130%2Bc%5C%5Cw%5E2%2B2%28w%29%283%29%2B%20%283%29%5E2%20%3D130%2B%283%29%5E2%5C%5CThe%5C%3Aleft%5C%3Aside%5C%3Abecomes%3A%20a%5E2%2B2ab%2Bb%5E2%5C%5CWe%5C%3Acan%5C%3Awrite%5C%3Ait%5C%3Aas%3A%20%28a%2Bb%29%5E2%5C%5CWe%5C%3Ahave%5C%3Aa%3Dw%5C%3A%20and%5C%3A%20b%3D3%5C%5C%28w%2B3%29%5E2%3D130%2B9%5C%5C%28w%2B3%29%5E2%3D139)
So, The value of c that will result in a perfect square trinomial is (3)^2 or 9
The perfect square trinomial is: ![(w+3)^2=139](https://tex.z-dn.net/?f=%28w%2B3%29%5E2%3D139)
Since the properties of a rhombus tells us that both pairs of opposite sides are congruent we can therefore conclude that DB will be 12
Find how many numbers are divisible by two and three, then place that number above the total number of items in the set as a fraction. Like this: