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Sveta_85 [38]
3 years ago
12

I'm desperate for help!!! Please!!!

Mathematics
1 answer:
amm18123 years ago
8 0
A. sqrt(15)/15    B. 4nsqrt(2n)    C. 7sqrt(7)    D. 3sqrt(2)

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1/4 + 1/3= {Khan Academy btw}
zmey [24]

Answer: \frac{7}{12}

Step-by-step explanation:

\frac{1}{4} +\frac{1}{3}=\frac{(1)(3)+(4)(1)}{(4)(3)}  =\frac{3+4}{12} =\frac{7}{12}

5 0
2 years ago
Suppose you have the following recursion formula a1 = 1, a2 = 2, and an = a(n - 1)+ a(n - 2)for integers n ≥ 3. How would you de
julia-pushkina [17]
A3=3
a4=5
a5=8

This sequence has properties similar to those of the Fibonacci sequence.
8 0
3 years ago
Fill in the table using this function rule. y=-3x-3 ​
pashok25 [27]

Answer:

When x = -2, y = 3

When x = -1, y = 0

When x = 0, y = -3

When x = 1, y = -6

Step-by-step explanation:

Given:

y = -3x - 3

Fill in the table using the following value for x

When x = -2

y = -3x - 3

y = -3(-2) - 3

y = 6 - 3

y = 3

When x = -2, y = 3

When x = -1

y = -3x - 3

y = -3(-1) - 3

y = 3 - 3

y = 0

When x = -1, y = 0

When x = 0

y = -3x - 3

y = -3(0) - 3

y = 0 - 3

y = -3

When x = 0, y = -3

When x = 1

y = -3x - 3

y = -3(1) - 3

y = -3 - 3

y = -6

When x = 1, y = -6

3 0
2 years ago
Totes need help on my question 3^-34x+5678-(-89÷32-2100035
Allushta [10]
<span><span><span>‌<span>1/16677181699666568</span></span>‌x</span>+<span>‌<span>67382905/<span>32</span></span></span></span>
5 0
3 years ago
Read 2 more answers
A positive integer from one to six is to be chosen by casting a die. Thus the elements c of the sample space C are 1, 2, 3, 4, 5
Alex_Xolod [135]

Answer with Step-by-step explanation:

We are given that six integers 1,2,3,4,5 and 6.

We are given that sample space

C={1,2,3,4,5,6}

Probability of each element=\frac{1}{6}

We have to find that P(C_1),P(C_2),P(C_1\cap C_2) \;and\; P(C_1\cup C_2)

Total number of elements=6

C_1={1,2,3,4}

Number of elements in C_1=4

P(E)=\frac{number\;of\;favorable \;cases}{Total;number \;of\;cases}

Using the formula

P(C_1)=\frac{4}{6}=\frac{2}{3}

C_2={3,4,5,6}

Number of elements in C_2=4

P(C_2)=\frac{4}{6}=\frac{2}{3}

C_1\cap C_2={3,4}

Number of elements in (C_1\cap C_2)=2

P(C_1\cap C_2)=\frac{2}{6}=\frac{1}{3}

C_1\cup C_2={1,2,3,4,5,6}

P(C_1\cup C_2)=\frac{6}{6}=1

4 0
3 years ago
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