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lorasvet [3.4K]

2 years ago

11

Kayla has 24 yellow beads and 36 green beads a. what is the greatest number of necklaces she could make? b. how many yellow bead

s would be in each necklace? c. how many green beads would be in each necklace?

1 answer:

creativ13 [48]2 years ago

7 0

She can make 6 necklaces and these will be 4 yellow and 6 green on each necklace

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-3(7w-3)+8w=-5(w+1) what is the value of w

bogdanovich [222]

Answer:

2

Step-by-step explanation:

Step 1:

- 21w + 9 + 8w = - 5w - 5

Step 2:

- 13w + 9 = - 5w - 5

Step 3:

- 7w + 9 = - 5

Step 4:

- 7w = - 14

Step 5:

14 = 7w

Answer:

w = 2

Hope This Helps :)

6 0

2 years ago

8, 2, 0, 2, 8, 18<br> What are the next three numbers in the sequence?<br> What is the pattern?

Tom [10]

Answer:

22,40,62

Step-by-step explanation:

0+2=2 2+6=8 8+10=18

2 6 10 14 18 22 four in between

so you will add these numbers each time

18+14=22 22+18=40 40+22=62

4 0

2 years ago

If cosθ=2/3 find cscθ using identities. This is in quad. 1.

laiz [17]

Step-by-step explanation:

cos θ = 2/3

$\quad \twoheadrightarrow\sf {cos \; \theta = \dfrac{Base}{Hypotenuse} } \\$

Hence, base = 2 units and hypotenuse = 3 units.

$\quad \twoheadrightarrow\sf { H^2 = B^2 + P^2} \\$

$\quad \twoheadrightarrow\sf { P^2 = H^2 - B^2} \\$

$\quad \twoheadrightarrow\sf { P^2 = (3)^2 - (2)^2} \\$

$\quad \twoheadrightarrow\sf { P^2 = 9- 4} \\$

$\quad \twoheadrightarrow\bf { P = \sqrt{5}} \\$

Now, we know that :

$\quad \twoheadrightarrow\sf {cosec \; \theta = \dfrac{Hypotenuse}{Perpendicular} } \\$

$\quad \twoheadrightarrow\bf{cosec \; \theta = \dfrac{3}{\sqrt{5}} } \\$

Therefore, the required answer is 3/√5.

4 0

1 year ago

In tossing four fair dice, what is the probability of tossing, at most, one 3

11Alexandr11 [23.1K]

<u>Answer- </u>

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

<u>Solution-</u>

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$ ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$ ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

7 0

2 years ago

PLEASE HELP ME WITH THIS! ASAP!!!!!!!

marusya05 [52]

Hold on i believe its C but let me go look on google and i'll get back to you to sure

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2 years ago