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shusha [124]
2 years ago
11

Mrs. Walker is cutting construction paper for an art project. She has 16 students, and 6 large pieces of construction paper. How

much paper will each student receive?
Mathematics
1 answer:
ikadub [295]2 years ago
8 0

Answer:  \frac{3}{8} pieces of paper each student will receive.

Step-by-step explanation:

Since we have given that

Number of pieces of construction paper = 6

Number of students = 16

According to question, each student will receive the construction paper for an art project.

So, Number of pieces of paper each student will receive is given by

\frac{6}{16}\\\\=\frac{3}{8}

Hence, \frac{3}{8} pieces of paper each student will receive.

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Sarah buys 2 candy bars for $1.45. how much would one candy bar cost
sattari [20]
2 = $1.45
1 = $0.725

anything to ask please pm me
4 0
2 years ago
PLEASE HELP ME!
algol13

Step-by-step explanation:

1.\sum_{i=1}^{5}3i

The simplest method is "brute force".  Calculate each term and add them up.

∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)

∑ = 3 + 6 + 9 + 12 + 15

∑ = 45

2.\sum_{k=1}^{4}(2k)^{2}

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²

∑ = 4 + 16 + 36 + 64

∑ = 120

3.\sum_{k=3}^{6}(2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)

∑ = -4 + -2 + 0 + 2

∑ = -4

4. 1 + 1/4 + 1/16 + 1/64 + 1/256

This is a geometric sequence where the first term is 1 and the common ratio is 1/4.  The nth term is:

a = 1 (1/4)ⁿ⁻¹

So the series is:

\sum_{j=1}^{7}(\frac{1}{4})^{j-1}

5. -5 + -1 + 3 + 7 + 11

This is an arithmetic sequence where the first term is -5 and the common difference is 4.  The nth term is:

a = -5 + 4(n−1)

a = -5 + 4n − 4

a = 4n − 9

So the series is:

\sum_{j=1}^{5}(4j-9)

5 0
2 years ago
An aquarium with a square base has no top. There is a metal frame. Glass costs 5 dollars/m2 and the frame costs 2 dollars/m. The
adell [148]

Answer:

C = 420/h + 400

Step-by-step explanation:

Let s be the side of the square base.

Let h be the height

Volume = s*h

20 = s*h

s = 20/h

Cost of glass is

5(20/h) + 5(4* h*20/h)

= 100/h + 400

Cost of frame is

2*4(20/h) + 2*4(20/h)

= 160/h + 160/h

= 320/h

Total cost = C

C = cost of glass + cost of frame

C = 100/h + 400 + 320/h

C = 420/h + 400

3 0
2 years ago
OMG PLEASE HELP ME ASAP!!!!!!! A local charity sponsors a 5K race to raise money. It receives $55 per race entry and $10,000 in
Dominik [7]
Start with how much profit they are making off each race entry. People pay $55 to race, but $15 of that is expenses so they are only profiting $40 for each entry. Now write one side of the equality. They start with $10,000 in donations, and then have a $40 profit for each race entry. So 10,000+40x. X will represent the unknown number of race entries. What do we want that expression to be equal to? We want 10000+40x>55000. It can also be greater than or equal to, not just greater than. Solve for x. Subtract 10000 from each side resulting in 40x>45000. Divide each side by 40 to solve for x. X>1125. X needs to bbe greater than or equal to 1125. If there are 1125 race entries, the charity will profit exactly $55000, so the lowest number of race entries is 1125
3 0
2 years ago
A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modele
tresset_1 [31]

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Step-by-step explanation:

\text{The equation for ball height} : 6+30\cdot t-16\cdot t^{2}\\\text{The equation for shot blocker's height} : 9+25\cdot t-16\cdot t^{2}

But, the shot is made before two tenths of a second or 0.2 seconds therefore modified equation for ball height is :

6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}

Now for the shot to be blocked,the height of shot blocker must be greater than the height of the ball which is shot before 0.2 seconds :

\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot (t-0.2)-16\cdot (t-0.2)^{2}\\\implies 9+25\cdot t-16\cdot t^{2}\geq 6+30\cdot t-6-16\cdot (t^{2}-0.4\cdot t+0.04)\\\implies9+25\cdot t-16\cdot t^{2}\geq 30\cdot t-16\cdot t^{2}+6.4\cdot t-0.64\\\implies 9+25\cdot t\geq 36.4\cdot t-0.64\\\implies 9.64\geq 11.4\cdot t\\\\\implies t\leq \frac{9.64}{11.4}\approx 0.846

Hence, the shot was blocked between 0.84 and 0.85 seconds after the shot is launched.

7 0
2 years ago
Read 2 more answers
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