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3 years ago

What is the domain and range of this graph?

Mathematics

1 answer:

Gala2k [10]3 years ago

5 0

Answer:

Domain is all real numbers. Range is [36- negative infinity]

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I need help plsss. im stuck

noname [10]

Answer:

h = 6 cm

Step-by-step explanation:

Volume of cylinder = 54 π cm³

πr²h = 54π

π * 3 * 3 * h = 54π

$h = \frac{54 \pi }{\pi *3*3}\\\\h= 6$

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2 years ago

ILEGIT NEED HELP BADDDDDDD

sergiy2304 [10]

Please restore my faith in humans and try to understand this concept

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2 years ago

a basketball team played five games . in those games , the team won by 7 points , lost by 3 , won by 4, lost by 2 , and won by 9

Paul [167]

Answer: 3

Step-by-step explanation:

7-3+4-2+9=15

15/5=3

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2 years ago

Help please i don’t know how to do it

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

2 years ago

I have nooooo clue, please help

Aleksandr-060686 [28]

Answer:

case a) $x^{2}=3y$ ----> open up

case b) $x^{2}=-10y$ ----> open down

case c) $y^{2}=-2x$ ----> open left

case d) $y^{2}=6x$ ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

$x=a(y-k)^{2}+h$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

$x^{2}=3y$

$y=(1/3)x^{2}$

$a=(1/3)$

$a>0$

therefore

The parabola open up

case b) we have

$x^{2}=-10y$

$y=-(1/10)x^{2}$

$a=-(1/10)$

$a$

therefore

The parabola open down

case c) we have

$y^{2}=-2x$

$x=-(1/2)y^{2}$

$a=-(1/2)$

$a$

therefore

The parabola open to the left

case d) we have

$y^{2}=6x$

$x=(1/6)y^{2}$

$a=(1/6)$

$a>0$

therefore

The parabola open to the right

3 0

2 years ago