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natima [27]
3 years ago
15

What is the answer ?

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0

Answer:

Step-by-step explanation:

718+200+40=718+240=958

Julli [10]3 years ago
4 0

Answer:

C. 718 + 240

Step-by-step explanation:

As you can see, the mark is first on 718. From there, they add 200 and then 40. 200 + 40 = 240

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Im not in expert at this but i did the best i could im 80% sure this is the anser if not im sorry    x > 30 
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What is (-6,-2)×1/2 ?​
Svetradugi [14.3K]

Answer:

the answer is (-4 )

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How do you solve the system of the linear equation by substitution?<br> y=x-4<br> 4x-y=3
Bingel [31]

Answer:

x=-1/3, y=-13/3. (-1/3, -13/3).

Step-by-step explanation:

y=x-4

4x-y=3

------------

4x-(x-4)=3

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y=-1/3-4

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3 years ago
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I need help.. i really want to go sleep.. thank you so much...
Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given  \sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}

To verify that the above equality is true or false:

Now find \sum\limits_{k=0}^8\frac{1}{k+3}

Expanding the summation we get

\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3} \sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

Now find \sum\limits_{i=3}^{11}\frac{1}{i}

Expanding the summation we get

\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}

 Comparing the two series  we get,

\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i} so the given equality is true.

2) Given \sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}

Verify the above equality is true or false

Now find \sum\limits_{k=0}^4\frac{3k+3}{k+6}

Expanding the summation we get

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}

\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}

now find \sum\limits_{i=1}^3\frac{3i}{i+5}

Expanding the summation we get

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}

\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}

Comparing the series we get that the given equality is false.

ie, \sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}

6 0
3 years ago
Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac
storchak [24]

Answer:

θ = πk, 1.318+2πk, 4.965+2πk

Step-by-step explanation:

4sin(2θ) − 2sin(θ) = 0

8sin(θ)cos(θ) - 2sin(θ) = 0

2sin(θ)[4cos(θ)-1] = 0

2sin(θ) = 0

sin(θ) = 0

θ = πk

4cos(θ)-1 = 0

4cos(θ) = 1

cos(θ) = 1/4

θ ≈ 1.318+2πk, 4.965+2πk

Therefore, θ = πk, 1.318+2πk, 4.965+2πk

4 0
2 years ago
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