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zhuklara [117]
3 years ago
10

What is the solution of the equation x2 + 8x + 5 = 0?

Mathematics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

x = -0.683 (3dp)

x = -7.317 (3dp)

Step-by-step explanation:

<u>Step 1: Apply the quadratic formula</u>

(-b±(\sqrt{b^{2}-4ac})) / 2a

a = 1 <em>(x²)</em>

b = 8 <em>(8x)</em>

c = 5 <em>(5)</em>

(-8±(\sqrt{8^{2}-4 x 1 x 5})) / 2 x 1

<u>Step 2: Simplify</u>

(-8±(\sqrt{44})) / 2

<u>Step 3: Solve</u>

You need to replace the ± with first a + and then a -.

(-8+(\sqrt{44})) / 2 = -0.683 (3dp) = x

(-8-(\sqrt{44})) / 2 = -7.317 (3dp) = x

Hope this helps!

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How is the series 5+11+17…+251 represented in summation notation?
NISA [10]

Answer:

\displaystyle \large{\sum_{n=1}^{42}(6n-1)

Step-by-step explanation:

Given:

  • Series 5+11+17+...+251

To find:

  • Summation notation of the given series

Summation Notation:

\displaystyle \large{\sum_{k=1}^n a_k}

Where n is the number of terms and \displaystyle \large{a_k} is general term.

First, determine what kind of series it is, there are two main series that everyone should know:

  • Arithmetic Series

A series that has common difference.

  • Geometric Series

A series that has common ratio.

If you notice and keep subtracting the next term with previous term:

  • 11-5 = 6
  • 17-11 = 6

Two common difference, we can in fact say that the series is arithmetic one. Since we know the type of series, we have to find the number of terms.

Now that brings us to arithmetic sequence, we know that first term is 5 and last term is 251, we’ll be finding both general term and number of term using arithmetic sequence:

<u>Arithmetic Sequence</u>

\displaystyle \large{a_n=a_1+(n-1)d}

Where \displaystyle \large{a_n} is the nth term, \displaystyle \large{a_1} is the first term and \displaystyle \large{d} is the common difference:

So for our general term:

\displaystyle \large{a_n=5+(n-1)6}\\\displaystyle \large{a_n=5+6n-6}\\\displaystyle \large{a_n=6n-1}

And for number of terms, substitute \displaystyle \large{a_n} = 251 and solve for n:

\displaystyle \large{251=6n-1}\\\displaystyle \large{252=6n}\\\displaystyle \large{n=42}

Now we can convert the series to summation notation as given the formula above, substitute as we get:

\displaystyle \large{\sum_{n=1}^{42}(6n-1)

5 0
2 years ago
2(8r+5)-3=4(4r-1)+11 is it an only solution or a no solution or infinite solution
pogonyaev

Answer:

Infinitely many solutions.

Step-by-step explanation:

Let's begin by carrying out the indicated multiplications, which must be done before any addition or subtraction:

2(8r+5)-3=4(4r-1)+11  becomes  16r + 10 - 3 = 16r - 4 + 11.

Subtracting 16r from both sides, we get 10 - 3 = - 4 + 11, or 7 = 7

This is always true, so we can conclude that this equation has infinitely many solutions.

3 0
2 years ago
Some students are going on a field trip. The cost of the trip is $100, in addition to $7per student. Write an equation that rela
grigory [225]

Answer: C= 100+7n

Step-by-step explanation:

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2 years ago
What property is shown by the equation?
Tom [10]
E. negative one property of multiplication.
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3 years ago
Let the correct answer
grigory [225]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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