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siniylev [52]
3 years ago
12

2xsquared-9x+1=0 value of discriminant and number of real solutions

Mathematics
1 answer:
shtirl [24]3 years ago
7 0

Answer:

• discriminant: 73

• # of real solutions: 2

Step-by-step explanation:

Comparing the equation ...

2x^2 -9x +1 = 0

to the generic form ...

ax^2 +bx +c = 0

we find the coefficient values to be ...

a = 2; b = -9; c = 1

That makes the value of the discriminant, (b^2 -4ac), be ...

(-9)^2 -4(2)(1) = 81 -8 = 73

Since the discriminant is positive, the number of real solutions is 2.

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I can’t figure out what the half circle means the answer options are 163,129,149 and 186
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3 years ago
Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of
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Answer:

Part 1) csc(\theta)=-\frac{\sqrt{85}}{7}

Part 2) sin(\theta)=-\frac{7}{\sqrt{85}}  or sin(\theta)=-\frac{7\sqrt{85}}{85}

Part 3) tan(\theta)=-\frac{7}{6}

Part 4) cos(\theta)=\frac{6}{\sqrt{85}} or cos(\theta)=\frac{6\sqrt{85}}{85}

Part 5) sec(\theta)=\frac{\sqrt{85}}{6}

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

so

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

1+cot^{2}(\theta)=csc^{2}(\theta)

we have

cot(\theta)=-\frac{6}{7}

substitute

1+(-\frac{6}{7})^{2}=csc^{2}(\theta)

1+\frac{36}{49}=csc^{2}(\theta)

\frac{85}{49}=csc^{2}(\theta)rewrite

csc(\theta)=-\frac{\sqrt{85}}{7} ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

csc(\theta)=\frac{1}{sin(\theta)}

we have

csc(\theta)=-\frac{\sqrt{85}}{7}

therefore

sin(\theta)=-\frac{7}{\sqrt{85}}

or

sin(\theta)=-\frac{7\sqrt{85}}{85}

step 3

Find the value of  tan(θ)

we know that

tan(\theta)=\frac{1}{cot(\theta)}

we have

cot(\theta)=-\frac{6}{7}

therefore

tan(\theta)=-\frac{7}{6}

step 4

Find the value of cos(θ)

we know that

sin^{2}(\theta)+cos^{2}(\theta)=1

we have

sin(\theta)=-\frac{7}{\sqrt{85}}

substitute

(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1

\frac{49}{85}+cos^{2}(\theta)=1

cos^{2}(\theta)=1-\frac{49}{85}

cos^{2}(\theta)=\frac{36}{85}

cos(\theta)=\frac{6}{\sqrt{85}} ------> the cosine is positive

or

cos(\theta)=\frac{6\sqrt{85}}{85}

step 5

Find the value of sec(θ)

we know that

sec(\theta)=\frac{1}{cos(\theta)}

we have

cos(\theta)=\frac{6}{\sqrt{85}}

therefore

sec(\theta)=\frac{\sqrt{85}}{6} ----> is positive

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4 years ago
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