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kow [346]
3 years ago
15

Evaluate 1/3y + 1/6 when y equals 1/2. then add the result by 1/6

Mathematics
1 answer:
tiny-mole [99]3 years ago
3 0
The answer is right below:

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Change to form fraction 35.8 % Give way
Radda [10]
Your answer is \frac{179}{5}
6 0
3 years ago
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In apirl is rained 1/4 of a inch on 15 diffret day how many inches of rain fell
Ne4ueva [31]

In 15 diffret days 15/4 inches of rain fell in April month if every day rained 1/4 of a inch .

How to find the rain fell in 15 diffret days ?

We know that rain in every day is 1/4 of a inch

So for 15 days

1/4+1/4+1/4+......15times\\=15/4

So in 15 diffret days in April rain fell is 15/4 inches.

Learn more about the rain fall is here :

brainly.com/question/28049911

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8 0
2 years ago
I NEED DESPERATE HELP BADDIES VAWYCVYAGVFYGQEVEF
gtnhenbr [62]

Answer:

BCD is the outer angle of triangle ABC

=> BCD = ABC + BAC

<=> 150⁰ = ABC + 118⁰

=> ABC = 150⁰ - 118⁰

<=> ABC = 32⁰

5 0
3 years ago
Read 2 more answers
Write an algebraic expression for the phrase five less than the product of six and N
GenaCL600 [577]

Answer:

6N - 5

Step-by-step explanation:

five less than the product of six and N: <em>6N</em>

five less than the product of six and N: <em>6N </em><em>- 5</em>

Answer: 6N - 5

6 0
3 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

4 0
4 years ago
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