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ruslelena [56]
3 years ago
6

When Earth, the moon, and the sun align, which order would lead to a solar eclipse?

Mathematics
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

Moon is between sun and earth

then:

the order is:

sun - moon - earth

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What are the exact solutions of x2 − 5x − 7 = 0, where x equals negative b plus or minus the square root of b squared minus 4 ti
mart [117]

Using The Almighty Formula (attached)

Note the symbol ^ is used to denote a raise to power

Where from the equation given x^2-5x-7

With a,b,c coefficients

a=1 b=-5 c=-7

From the Almighty Formula

√b^2-4ac=√(-5)^2-4(-7)=√53

Substituting in the Almighty Formula

-(-5) ±√53/2(1)= 5±√53/2

x=5+√53/2 and x=5-√53/2

6 0
3 years ago
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I only need help on question Q9 (a)
Maslowich
Not sure if this is it
The parts shaded represent 4/10, where 10 represents 360 degrees because that is the entire circle, therefore 40% of 360= 144 degrees
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3 years ago
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<img src="https://tex.z-dn.net/?f=The%20%5C%3A%20%20value%20%20%5C%3A%20of%20%20%5C%3A%20%20%5Csqrt%7B6%20%2B%20%20%5Csqrt%7B6%2
saul85 [17]

\large\underline{\sf{Solution-}}

<u>Let us assume that:</u>

\sf \longmapsto x =  \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ... \infty } } }

We can also write it as:

\sf \longmapsto x =  \sqrt{6 +  x }

Squaring both sides, we get:

\sf \longmapsto  {x}^{2}  =6 +  x

\sf \longmapsto  {x}^{2} - x - 6  =0

By splitting the middle term:

\sf \longmapsto  {x}^{2} - 3x + 2x - 6  =0

\sf \longmapsto x(x - 3) + 2(x - 3 ) =0

\sf \longmapsto (x+ 2)(x - 3 ) =0

<u>Therefore:</u>

\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}

\sf \longmapsto x =  - 2,3

<u>But x cannot be negative. </u>

\sf \longmapsto x = 3

Therefore, the value of the expression is 3.

\large\underline{\sf{Verification-}}

Given:

\sf\longmapsto x=3

We can also write it as:

\sf\longmapsto x = \sqrt{9}

\sf\longmapsto x = \sqrt{6+3}

\sf\longmapsto x = \sqrt{6 + \sqrt{9}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}

This pattern will continue.

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}

7 0
3 years ago
Can someone help me please ?
Alex Ar [27]

Answer:

V= 40.9

Step-by-step explanation:

the top and the bottom side both equal 26.2, if you multiply that number by 2 (Because 2 sides have that same number) you get 52.4. The next step is to take the total perimeter which is 134.2, and subtract 52.4 which brings you to 81.8. Then you divide 81.8 by 2 to get your final answer of 40.9, which is the value of V. Hope this helps!

6 0
3 years ago
a cell phone company plans to market a new smartphone. they have already sold 612 units durning the first week of the campaign.
Vadim26 [7]

The first term is 612.

The common ratio is 1.08 and

The recursive rule is a_{n} = a^{n-1} \times r

<u>Step-by-step explanation:</u>

the question to the problem is to write the values of the first term, common ratio, and expression for the recursive rule.

<u>The first term :</u>

In geometric sequence, the first term is given as a_{1}.

⇒ a_{1} = 612

Now, the geometric sequence follows as 612, 661, ........

<u>The common ratio (r) :</u>

It is the ratio between two consecutive numbers in the sequence.

Therefore, to determine the common ratio, you just divide the number from the number preceding it in the sequence.

⇒ r = 661 divided by 612

⇒ r = 1.08

<u>To find the recursive rule :</u>

A geometric series is of the form  a,ar,ar2,ar3,ar4,ar5........

Here, first term a_{1} = a and other terms are obtained by multiplying by r.

  • Observe that each term is r times the previous term.
  • Hence to get nth term we multiply (n−1)th term by r .

The recursive rule is of the form a_{n} = a^{n-1} \times r

This is called recursive formula for geometric sequence.

We know that r = 1.08 and a_{1} = 612.

To find the second term a_{2}, use the recursive rule a_{n} = a^{n-1} \times r

⇒ a_{2} = a^{2-1}\times r

⇒ a_{2} = a^{1}\times r

⇒ a_{2} = 612\times 1.08

⇒ a_{2} = 661

3 0
3 years ago
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