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NemiM [27]
3 years ago
6

Solve the equation below: 5x^2 - 55 = 90

Mathematics
1 answer:
Leno4ka [110]3 years ago
3 0
X= √29, -√29

The answer to this question would be ^
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Solve the equation: 5x -5
Zanzabum

Answer: 5(x-1)

Step-by-step explanation:

4 0
3 years ago
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5<br>3х - 10<br>15. х - 3x —<br>10<br>x-(3x -<br>3​
Levart [38]

Answer:

3 - 151 x

Step-by-step explanation:

d/dx(5×3 x - 10×15 x - 3 x - 10 x - (3 x - 3)) = -151

Indefinite integral:

integral(3 - 151 x) dx = 3 x - 75.5 x^2 + constant

Definite integral after subtraction of diverging parts:

integral_0^∞ ((3 - 151 x) - (3 - 151 x)) dx = 0

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2 years ago
at the grocery sale you want to buy 4 and 9 tenths lb of cheese what decimal should the digital scale show
Alexxx [7]

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4.9 lb

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3 years ago
 Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III
Paha777 [63]

Given:

\sin x=-\dfrac{15}{17}

x lies in the III quadrant.

To find:

The values of \sin 2x, \cos 2x, \tan 2x.

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others  are negative.

We know that,

\sin^2 x+\cos^2 x=1

(-\dfrac{15}{17})^2+\cos^2 x=1

\cos^2 x=1-\dfrac{225}{289}

\cos x=\pm\sqrt{\dfrac{289-225}{289}}

x lies in the III quadrant. So,

\cos x=-\sqrt{\dfrac{64}{289}}

\cos x=-\dfrac{8}{17}

Now,

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

\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})

\sin 2x=-\dfrac{240}{289}

We know that,

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

\cos 2x=1-2(-\dfrac{15}{17})^2

\cos 2x=1-2(\dfrac{225}{289})

\cos 2x=\dfrac{289-450}{289}

\cos 2x=-\dfrac{161}{289}

Using the trigonometric ratios, we get

\tan 2x=\dfrac{\sin 2x}{\cos 2x}

\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}

\tan 2x=\dfrac{240}{161}

Hence, the required values are \sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}.

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3 years ago
A line perpendicular to y=-3x+6 graph it
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