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4 years ago

Which statement about the following equation is true?

Mathematics

2 answers:

Leviafan [203]4 years ago

6 0

1. 3x^2-8x+5=5x^2
2x^2+8x-5

Discriminant = 64+40 = 104 > 0
The answer is 2

Makovka662 [10]4 years ago

3 0

Which statement about the following equation is true?

3x^2-8x+5=5x^2


2) The discriminant is greater than 0, so there are two real roots.

Let's solve your equation step-by-step.3x2−8x+5=5x2Step 1: Subtract 5x^2 from both sides.3x2−8x+5−5x2=5x2−5x2−2x2−8x+5=0Step 2: Use quadratic formula with a=-2, b=-8, c=5.x=−b√±b2−4ac/2a
x=−(−8)±√(−8)2−4(−2)(5)/2(−2)
x=8±√104/−4
x=−2+−1/2√26 or x=−2+1/2√26

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3 years ago

Read 2 more answers

Given that sec (x) = 2 and cosec (x) is negative,

weeeeeb [17]

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i) sin(2x) = $-\frac{\sqrt{3}}{2}$

ii) cot(x+360) = $-\frac{\sqrt{3}}{3}$

iii) sin(x-180) = $\frac{\sqrt{3}}{2}$

Step-by-step explanation:

sec(x) = 2

Since cos(x) is reciprocal of sec(x), this means:

cos(x) = $\frac{1}{2}$

cosec(x) is negative , this means sin(x) is also negative. The only quadrant where cos(x), sec(x) are positive and sin(x), cosec(x) are negative is the 4th quadrant. Hence the terminal arm of the angle x is in 4th quadrant.

Part i)

sin(2x) can be simplified as:

sin(2x) = 2 sin(x) cos(x)

First we need to find the value of sin(x). According to Pythagorean identity:

$sin^{2}(x)=1-cos^{2}(x)\\\\ sin(x)=\pm \sqrt{1-cos^{2}(x)}$

Since, angle is in 4th quadrant, sin(x) will be negative. So considering the negative value of sin(x) and substituting the value of cos(x), we get:

$sin(x)=- \sqrt{1-cos^{2}(x)}\\\\ sin(x)=-\sqrt{1-(\frac{1}{2})^{2}}\\\\ sin(x)=-\frac{\sqrt{3}}{2}$

So,

$sin(2x)=2 \times -\frac{\sqrt{3} }{2} \times \frac{1}{2}\\\\ sin(2x)=-\frac{\sqrt{3}}{2}$

Part ii)

We have to find cot(x + 360)

An addition of 360 degrees to the angle brings it back to the same terminal point. So the trigonometric ratios of the original angle and new angle after adding 360 or any multiple of 360 stay the same. i.e.

cot(x + 360) = cot(x)

$cot(x) = \frac{cos(x)}{sin(x)}\\$

Using the values, we get:

$cot(x)=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2} }\\\\ cot(x)=-\frac{\sqrt{3}}{3}$

Part iii)

We need to find the value of sin(x - 180)

sin(x - 180) = - sin(x)

Addition or subtraction of 180 degrees changes the angle by 2 quadrants. The sign of sin(x) becomes opposite if the angle jumps by 2 quadrants. For example, sin(x) is positive in 1st quadrant and negative in 3rd quadrant.

So,

sin(x - 180) = $-(-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2}$