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Nikolay [14]
3 years ago
9

Find the discriminant of the quadratic $3x^2 - 7x + 6.$

Mathematics
2 answers:
mariarad [96]3 years ago
6 0

Answer:

Discriminant of the quadratic is-23

Step-by-step explanation:

The given quadratic function is 3x^2-7x+6

Comparing with the expression ax^2+bx+c

a = 3, b = -7, c = 6

The discriminant of the quadratic is given by D=b^2-4ac

Substituting the known values, discriminant of the quadratic is

D=(-7)^2-4(3)(6)\\\\D=49-72\\\\D=-23

Therefore, discriminant of the quadratic is-23

Studentka2010 [4]3 years ago
4 0
Solve for x:
3 x^2 - 7 x + 6 = 0

Divide both sides by 3:
x^2 - (7 x)/3 + 2 = 0

Subtract 2 from both sides:
x^2 - (7 x)/3 = -2

Add 49/36 to both sides:

x^2 - (7 x)/3 + 49/36 = -23/36

Write the left hand side as a square:
(x - 7/6)^2 = -23/36

Take the square root of both sides:
x - 7/6 = (i sqrt(23))/6 or x - 7/6 = -(i sqrt(23))/6

Add 7/6 to both sides:
x = 7/6 + (i sqrt(23))/6 or x - 7/6 = -(i sqrt(23))/6

Add 7/6 to both sides:

Answer: x = 7/6 + (i sqrt(23))/6 or x = 7/6 - (i sqrt(23))/6
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At Dorthys hair salon, 4 gallons of shampoo will last 3 weeks. She has 7 gallons of shampoo. How much more must she order if she
lara31 [8.8K]
X=number of gallons that she must order.

4 gallons-----------------3 weeks
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5 0
3 years ago
If the price of pineapple juice was $4.50 a gallon and it is now $5.75 a gallon, what is the percentage change in price
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Step-by-step explanation:

5 0
2 years ago
Rewrite with only sin x and cos x.
Annette [7]

Option A

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

cos 3x = cos(x + 2x)

We know that,

\cos (a+b)=\cos a \cos b-\sin a \sin b

Therefore,

\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x  ---- eqn 1

We know that,

\sin 2 x=2 \sin x \cos x

\cos 2 x=\cos ^{2} x-\sin ^{2} x

Substituting these values in eqn 1

\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)  -------- eqn 2

We know that,

\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x

Applying this in above eqn 2, we get

\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)

\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}

\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x

Therefore,

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

Option A is correct

7 0
3 years ago
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attashe74 [19]
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5 0
3 years ago
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