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valentina_108 [34]

2 years ago

9

A half of a class and a half of a student got A on a test. If yet another student got an A, then two-thirds of the class would h

ave an A for the test! How many students are there in that class? Please Help!

1 answer:

Cerrena [4.2K]2 years ago

6 0

Answer:

21

Step-by-step explanation:

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Solve for X kinda confused i don’t know what to do!!

sweet [91]

Answer:

16.9265381881

Step-by-step explanation:

Tangent = Opposite/Adjacent.

Thus, tan62 = x/9 --> tan62 = 1.88072646535, multiply by 9 to get x, 1.88072646535* 9 = 16.9265381881

7 0

2 years ago

Did I get this right?

wel

Yes hope it helps!!!!!!!!

4 0

2 years ago

What is the distance between the origin and the point (2,-5) on a graph? The options are: square root of 7. Square root of 29. 7

Sunny_sXe [5.5K]

Answer:

Square root of 29.

Step-by-step explanation:

Distance = sqrt(2^2 + (-5)^2)

= sqrt 29.

3 0

2 years ago

10) Use the equation x2 + 8x = 9 to answer the following questions (1 point for each part, 2 points total). .

GREYUIT [131]

Answer:

i might be wrong but i think its 10

Step-by-step explanation:

x2+8x=9

collet terms

10x=9

divide both side by 10

x = 9/10

alternative form

x = 0.9

8 0

2 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

2 years ago