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3 years ago

Zack watered his garden with 1 3/8

Mathematics

1 answer:

victus00 [196]3 years ago

3 0

Answer:

<u>The correct answer is that Zack used 2 7/8 gallons to water his garden on the fifth week.</u>

Step-by-step explanation:

Let's find out the pattern Zack is using to water his garden, this way:

First week = 1 3/8 gallons of water

Second week = 1 3/4 gallons of water = 1 3/8 + 3/8 = 1 6/8 = 1 3/4

Third week = 2 1/8 gallons of water = 1 3/4 + 3/8 = 1 6/8 + 3/8 = 2 1/8

The constant of adding every week is 3/8 gallons, therefore,

Fourth week = 2 1/8 + 3/8 = 2 4/8 = 2 1/2 gallons of water

Fifth week = 2 1/2 + 3/8 = 2 4/8 + 3/8 = 2 7/8 gallons of water

<u>The correct answer is that Zack used 2 7/8 gallons to water his garden on the fifth week.</u>

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Every year the Lee family has a big family dinner, and everyone brings a mug to exchange. The mugs are put on wooden hangers on

RideAnS [48]

Answer:

21 more mugs were brought this year than last year.

Step-by-step explanation:

We can solve by subtracting the initial amount of mugs from the final amount of mugs.

To find the amount of mugs hung up this year, we will use the equation: m = 12w where m = the number of mugs & w = the number of wooden hangers. The 12 represents the amount of mugs per wooden hanger.

We can plug in 3 for w since 3 wooden hangers were filled up and we would get 12(3) = m = 36 mugs

Now we subtract last year's total mugs hung up from this year's total mugs hung up.

We end up with 36 - 15 = 21

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3 years ago

Answer these plssssss

blondinia [14]

Answer:

1. 8/24

2. 9/27

3. 5/15

4. 10/30

5. 2/6

6.1/3

Step-by-step explanation:

Start with 1/3 then multiply that by the number that is given.

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3 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

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3 years ago

After t seconds, a ball in the air from the ground level reaches a height of h feet given by the equation h= 144t-16t^2 . After

Effectus [21]

H=0
Therefore:
144t-16t²=0
t (144-16t)=0
We have two equations:
1)
t=0 (when the ball is released)
2)
144-16t=0
-16t=-144
t=-144 / -16=9 (this is the time )

Answer: the ball hit the ground in 9 s.

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3 years ago

The ransin sports company has noted that the size of individual customer orders is normally distributed with a mean of $112 and

Sonja [21]

Standard error of the mean is computed by:

Standard error = SD/ sqrt N

Where:

N is the sample size

SD is the standard deviation

To get the standard deviation, you need to get the sqrt of the variance = sqrt 9 = 3

So plugging in our data:

Standard error = 3 / sqrt (16)

= 0.75

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3 years ago