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8_murik_8 [283]
2 years ago
11

What are the zeros of the quadratic function f(x) = 8x2 – 16x – 15?

Mathematics
2 answers:
marusya05 [52]2 years ago
4 0

8x^2 - 16x - 15 = 0

x = [ -(-16) +/- sqrt((-16)^2 - 4*8*-15) / ] 2*8

= 2.696, -0.696

denis23 [38]2 years ago
3 0

Answer:

The zeroes of the given polynomial f(x) are

x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}

Step-by-step explanation:

Given : Quadratic function f(x)=8x^2-16x-15

To find : What are the zeros of the quadratic function?

Solution :

Using quadratic formula of the general equation ax^2+bx+c=0 to get the roots is x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Comparing the given quadratic equation,

a = 8, b = -16  and  c = - 15.

Substitute the value in the formula,

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\\Rightarrow x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4\times 8\times (-15)}}{2\times 8}\\\\\\\Rightarrow x=\dfrac{16\pm\sqrt{256+480}}{16}\\\\\\\Rightarrow x=\dfrac{16\pm\sqrt{786}}{16}\\\\\\\Rightarrow x=\dfrac{16\pm4\sqrt{46}}{16}\\\\\\\Rightarrow x=\dfrac{4\pm\sqrt{46}}{4}\\\\\\\Rightarrow x=1\pm\sqrt{\dfrac{46}{16}}\\\\\\\Rightarrow x=1\pm\sqrt{\dfrac{23}{8}}\\\\\\\Rightarrow x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}.

Therefore, The zeroes of the given polynomial f(x) are

x=1-\sqrt{\dfrac{23}{8}},~~~x=1+\sqrt{\dfrac{23}{8}}

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When solving negative one over five -1/5 (x − 25) = 7, what is the correct sequence of operations
aleksandr82 [10.1K]

The solution would be like this for this specific problem:

 

\left( { - 5} \right)
\cdot \left( { - \frac{1}{5}} \right) \cdot \left( {x - 25} \right) = 7 \cdot
\left( { - 5} \right)

 

 

\left( { - 5} \right)
\cdot \left( { - \frac{1}{5}} \right) = 1

 

The correct sequence of operations when solving negative one over five (x − 25) = 7 would be multiply each side by −5, and adding 25 to each side.

5 0
2 years ago
Which point shows the location of 5 – 2i on the complex plane below? On a coordinate plane, points A, B, C, and D are shown. Poi
Romashka-Z-Leto [24]

Answer:

The Point C shows the location of 5-2i in the complex plane: 5 points to the right of the origin and 2 points down from the origin.

Step-by-step explanation:

We have the complex number 5-2i and we have to show the location of the point that represents that number in the complex plane

In the complex plane the real numbers are located in the horizontal axis, increasing to the right. The positives real numbers are at the right of the origin and the negatives to the left.

The complex numbers are located in the vertical axis, with the positives over the origin and the negatives below the origin.

This complex number 5-2i is the sum of a real part (5) and a imaginary part (-2i), so the point will be 5 units rigth on the horizontal axis (for the real part) and 2 units down in the vertical axis (for the imaginary part).

8 0
2 years ago
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Tony bought a $48 sweatshirt and used a coupon for a 10% discount. Keith bought an identical sweatshirt at a different store for
Westkost [7]
<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ final = original x multiplier

final = 48 x 0.9

final = 43.2

Tony paid $43.20

Subtract the two values:

43.2 - 42.95 = 0.25

Therefore, the correct option would be C. Keith paid $0.25 less than Tony paid

<h3><u>✽</u></h3>

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3 0
2 years ago
In ΔFGH, f = 37 cm, ∠F=28° and ∠G=124°. Find the length of g, to the nearest 10th of a centimeter.
MatroZZZ [7]

Answer:

  65.3 cm

Step-by-step explanation:

The law of sines can be used for this.

  g/sin(G) = f/sin(F)

  g = (37 cm)sin(124°)/sin(28°) ≈ 65.3 cm . . . . multiply by sin(G)

The length of g is about 65.3 cm.

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2 years ago
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gulaghasi [49]
The answer to your question is x = 8
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