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blsea [12.9K]
3 years ago
5

Please help me find x. I really don't get it.

Mathematics
1 answer:
inysia [295]3 years ago
3 0

Answer:

  x = 11

Step-by-step explanation:

The relationship between the sine and cosine functions can be written as ...

  sin(x) = cos(90 -x)

  sin(A) = cos(90 -A) = cos(B) . . . . substituting the given values

Equating arguments of the cosine function, we have ...

 90 -(3x+4) = 8x -35

  86 -3x = 8x -35

  86 +35 = 8x +3x . . . . . add 3x+35 to both sides

  121 = 11x . . . . . . . . . . . . collect terms

  121/11 = x = 11 . . . . . . . . divide by 11

_____

<em>Comment on the solution</em>

There are other applicable relationships between sine and cosine as well. The result is that there are many solutions to this equation. One set is ...

  11 +(32 8/11)k . . . for any integer k

Another set is ...

  61.8 +72k . . . . . for any integer k

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Find Sn for the arithmetic series 5+7+9 + … and determine the value of n for which the series has sum 165.
SVETLANKA909090 [29]

Answer:

see explanation

Step-by-step explanation:

the sum to n terms of an arithmetic sequence is

S_{n} = \frac{n}{2}[2a + (n - 1)d ]

where d is the common difference and a is the first term

here d = 9 - 7 = 7 - 5 = 2 and a = 5, hence

S_{n} = \frac{n}{2}[(2 × 5) + 2(n - 1) ]

                        = \frac{n}{2}(10 + 2n - 2)

                        = \frac{n}{2}(2n + 8)

                        = n² + 4n

When sum = 165, then

n² + 4n = 165 ← rearrange into standard form

n² + 4n - 165 = 0 ← in standard form

(n + 15)(n - 11) = 0 ← in factored form

equate each factor to zero and solve for n

n + 15 = 0 ⇒ n = - 15

n - 11 = 0 ⇒ n = 11

but n > 0 ⇒ n = 11



4 0
3 years ago
Solve for x. 4 In (4x) = 16
sveticcg [70]

Answer:

x= 4

Step-by-step explanation:

4 x 4= 16

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Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

8 0
2 years ago
Use the distributive property to expand -4 (-2/5 + 3 x). Which is an equivalent expression?
Brrunno [24]

Answer:

8/5-12x.

Step-by-step explanation:

To distribute, you simply need to multiply each of the terms inside of the parenthesis by -4. This will give you:

=-4(\frac{-2}{5})-4(3x)

=8/5-12x. This is your answer!

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