Solve the quadratic equation by factoring. 7K^2-6k+3=3

2 answers:

ivann1987 [24]3 years ago

6 0

Subtract 3 to get:

7k^2-6k=0

Clearly, we can factor out a k to find that:

k(7k-6)=0

If either k or 7k-6 is zero, the whole expression is zero, so our solutions are zero and 6/7.

Ilya [14]3 years ago

6 0

We make one side zero
subtract 3 from both sides
7k^2-6k=0
factor out k
(k)(7k-6)=0
set each to zero
k=0

7k-6=0
add 6 to both sides
7k=6
diivde by 7
k=6/7

k=0 or 6/7

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2 years ago

Hello I need help Finding the probability

Stolb23 [73]

Each petal of the region $R$ is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length $5\sqrt2$ , which implies the segment is subtended by an angle of $\dfrac\pi2$ . This means the area of the segment is

$\text{area}_{\text{segment}}=\text{area}_{\text{sector}}-\text{area}_{\text{triangle}}$
$\text{area}_{\text{segment}}=\dfrac{25\pi}4-\dfrac{25}2$

This means the area of one petal is $\dfrac{25\pi}2-25$ , and the area of $R$ is four times this, or $50\pi-100$ .

Meanwhile, the area of $G$ is simply the area of the square minus the area of $R$ , or $10^2-(50\pi-100)=200-50\pi$ .

So

$\mathbb P(X=R)=\dfrac{50\pi-100}{100}=\dfrac\pi2-1$
$\mathbb P(X=G)=\dfrac{200-50\pi}{100}=2-\dfrac\pi2$
$\mathbb P((X=R)\land(X=G))=0$ (provided these regions are indeed disjoint; it's hard to tell from the picture)
$\mathbb P((X=R)\lor(X=G))=\mathbb P(X=R)+\mathbb P(X=G)=1$