Determine which relation is a function.

Jordan bought StartFraction 9 Over 16 EndFraction of a pound of sunflower seeds and divided them evenly into 9 snack bags for hi

Mars2501 [29]

Answer: 1/16 of a pound

Step-by-step explanation:

Here is the complete question:

Jordan bought 9/16 of a pound of sunflower seeds and divided them evenly into 9 snack bags for his lunch. What is the weight of sunflower seeds in each snack bag?

1/16 of a pound

1/9of a pound

16/81 of a pound

81/16 pounds

From the question, we are told that Jordan bought 9/16 of a pound of sunflower seeds and then divided them equally into 9 snack bags for his lunch. To get the weight of sunflower seeds in each snack bag, we will divide 9/16 by 9 snack bags. This can be written mathematically as:

= 9/16 ÷ 9

= 9/16 × 1/9

= 1/16

The weight of sunflower seeds in each snack bag will be 1/16 of a pound.

3 0

4 years ago

Read 2 more answers

What is the integral of a square root? Please explain.

IrinaK [193]

If y = xⁿ

∫y dx = xⁿ⁺¹ / (n + 1) + C Provided n ≠ -1.

y = √x

y = x^(0.5)

∫y dx = x^(0.5+1) / (0.5 + 1) = x^(1.5) / 1.5 = x^(1.5) / (3/2)

∫y dx = (2/3) x^(1.5) + C.

∫y dx = (2/3) x^(3/2) + C.

∫y dx = (2/3)√x³ + C

5 0

3 years ago

Read 2 more answers

What do we mean by 16 1/4

zhenek [66]

Evaluate?

What’s the full question?

8 0

3 years ago

Read 2 more answers

Find the fourth point of a parallelogram given the first three (more than one correct answer is possible): (0,6), (5, -1) and (3

inysia [295]

Answer:

The fourth point of the parallelogram is one point among (-2,2), (2,10) and (8,-12).

Step-by-step explanation:

Given information: The first three vertices of parallelogram are (0,6), (5, -1) and (3,-5).

Let fourth point of the parallelogram is (a,b).

Diagonals of a parallelogram bisect each other. It means midpoint of both diagonals are same.

Midpoint formula:

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Case 1: If the point (0,6), (5, -1) and (3,-5) are consecutive, then pairs of opposite vertices are (0,6) and (3,-5), (5,-1) and (a,b).

$(\frac{0+3}{2},\frac{6-5}{2})=(\frac{5+a}{2},\frac{-1+b}{2})$

$(\frac{3}{2},\frac{1}{2})=(\frac{5+a}{2},\frac{-1+b}{2})$

On comparing both sides, we get

$\frac{3}{2}=\frac{5+a}{2}$

$3=5+a$

$a=-2$

$\frac{1}{2}=\frac{-1+b}{2}$

$1=-1+b$

$b=2$

It means the fourth point of the parallelogram is (-2,2).

Case 2: If the point (0,6), (5, -1) and (3,-5) are not consecutive, then pairs of opposite vertices are (0,6) and (5,-1), (3,-5) and (a,b).

$(\frac{0+5}{2},\frac{6-1}{2})=(\frac{3+a}{2},\frac{-5+b}{2})$

$(\frac{5}{2},\frac{5}{2})=(\frac{3+a}{2},\frac{-5+b}{2})$

On comparing both sides, we get

$\frac{5}{2}=\frac{3+a}{2}$

$5=3+a$

$a=2$

$\frac{5}{2}=\frac{-5+b}{2}$

$5=-5+b$

$b=10$

It means the fourth point of the parallelogram is (2,10).

Case 3: If the point (0,6), (5, -1) and (3,-5) are not consecutive, then pairs of opposite vertices are (5,-1) and (3,-5), (0,6) and (a,b).

$(\frac{5+3}{2},\frac{-1-5}{2})=(\frac{0+a}{2},\frac{6+b}{2})$

$(\frac{8}{2},\frac{-6}{2})=(\frac{a}{2},\frac{6+b}{2})$

On comparing both sides, we get

$\frac{8}{2}=\frac{a}{2}$

$8=a$

$\frac{-6}{2}=\frac{6+b}{2}$

$-6=6+b$

$b=-12$

It means the fourth point of the parallelogram is (8,-12).

5 0

4 years ago

Use the approach in Gauss's Problem to find the following sums of arithmetic

Agata [3.3K]

a. Let S be the first sum,

S = 1 + 2 + 3 + … + 97 + 98 + 99

If we reverse the order of terms, the value of the sum is unchanged:

S = 99 + 98 + 97 + … + 3 + 2 + 1

If we add up the terms in both version of S in the same positions, we end up adding 99 copies of quantities that sum to 100 :

S + S = (1 + 99) + (2 + 98) + … + (98 + 2) + (99 + 1)

2S = 100 + 100 + … + 100 + 100

2S = 99 × 100

S = (99 × 100)/2

Then S has a value of

S = 99 × 50

S = 4950

Aside: Suppose we had n terms in the sum, where n is some arbitrary positive integer. Call this sum ∑(n) (capital sigma). If ∑ is a sum of n terms, and we do the same manipulation as above, we would end up with

2 ∑(n) = n × (n + 1) ⇒ ∑(n) = n (n + 1)/2

b. Let S' be the second sum. It looks a lot like S, but the even numbers are missing. Let's put them back, but also include their negatives so the value of S' is unchanged. In doing so, we have

S' = 1 + 3 + 5 + … + 1001

S' = (1 + 2 + 3 + 4 + 5 + … + 1000 + 1001) - (2 + 4 + … + 1000)

The first group of terms is exactly the sum ∑(1001). Each term in the second grouped sum has a common factor of 2, which we can pull out to get

2 (1 + 2 + … + 500)

so this other group is also a function of ∑(500), and so

S' = ∑(10001) - 2 ∑(500) = 251,001

However, we want to use Gauss' method. We have a sum of the first 501 odd integers. (How do we know there 501? Starting with k = 1, any odd integer can be written as 2k - 1. Solve for k such that 2k - 1 = 1001.)

S' = 1 + 3 + 5 + … + 997 + 999 + 1001

S' = 1001 + 999 + 997 + … + 5 + 3 + 1

2S' = 501 × 1002

S' = 251,001

c/d. I think I've demonstrated enough of Gauss' approach for you to fill in the blanks yourself. To confirm the values you find, you should have

3 + 6 + 9 + … + 300 = 3 (1 + 2 + 3 + … + 100) = 3 ∑(100) = 15,150

and

4 + 8 + 12 + … + 400 = 4 (1 + 2 + 3 + … + 100) = 4 ∑(100) = 20,200

3 0

2 years ago