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3 years ago

PLEASE HELP ME YALL I NEED YOUR HELP LIKE A LOT. PLEASEEEE I DONT UNDERSTAND NOTHING I AM VERY DUMB

Mathematics

1 answer:

astraxan [27]3 years ago

5 0

(>_ means greater than or equal to, and <_ means less than or equal to)

Answer:

9: Domain: 0<_x<_10

10: Range: 0<_y<_32

11: Domain: 0<_x<_∞ Range: 10<_y<_∞

12: Domain: 0<_x<_∞ Range: y=25

13: Domain: 0<_x<_15 Range: 0<_y<_15

Step-by-step explanation:

Just look at the highest and lowest points for the x axis and the y axis, and that's your domain and range respectively

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nasty-shy [4]

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Step-by-step explanation:

The volume of a prism =Base Area× Height.

6 * 9 * 8 =

54 * 8 =

432 cm³

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1 year ago

Find x and y with steps

LuckyWell [14K]

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x=15, y=12

Step-by-step explanation:

Please see the attached pictures for full solution.

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2 years ago

The area of Tracy's backyard is 1 1/3 acres she plants a garden that take up 1/3 of the backyard what area in acres of the garde

attashe74 [19]

Answer:

$\frac{4}{9}\ acres$

Step-by-step explanation:

step 1

we know that

The area of Tracy's backyard is $1\frac{1}{3}\ acres$

Convert to an improper fraction

$1\frac{1}{3}=\frac{3*1+1}{3}=\frac{4}{3}\ acres$

step 2

To find the area of the garden, multiply the area of the backyard by 1/3

$\frac{4}{3}(\frac{1}{3})=\frac{4}{9}\ acres$

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2 years ago

Riddle me this what goes in the water dry and comes out dry

bulgar [2K]

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This is a guess but is it : A Shadow ?

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2 years ago

"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e

Art [367]

Answer:

a. v(t)= -6.78 $e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= $e^{\int\limits^ {}k \, dt }$ = $e^{kt}$ . We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$ v' + k $e^{kt}$ v = g $e^{kt}$ ⇒ [v $e^{kt}$ ]' = g $e^{kt}$ . Integrating, we have

∫ [v $e^{kt}$ ]' = ∫g $e^{kt}$

v $e^{kt}$ = $\frac{g}{k}$ $e^{kt}$ + c

v(t)= $\frac{g}{k}$ + c $e^{-kt}$ .

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c $e^{-16.33 * 0}$ = 16.33 + c

c = 9.55 -16.33 = -6.78.

So, v(t)= 16.33 - 6.78 $e^{-16.33t}$ . m/s = - 6.78 $e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78 $e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

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2 years ago

PLEASE HELP ME YALL I NEED YOUR HELP LIKE A LOT. PLEASEEEE I DONT UNDERSTAND NOTHING I AM VERY DUMB ​

PLEASE HELP ME YALL I NEED YOUR HELP LIKE A LOT. PLEASEEEE I DONT UNDERSTAND NOTHING I AM VERY DUMB