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2 years ago

9

The grass in Jamie’s yard grew 16 centimeters in 10 days. Kit was growing at a constant rate how many days did it take the grass

to grow 1centimeter

1 answer:

lapo4ka [179]2 years ago

7 0

0.625 day to grow 1 cm.

16/10 = 1.6 cm in day.

I hope that help ✌️

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Simplify<br> CLEAR Step By Step Explanation Please. <br> Will Mark Brainliest

siniylev [52]

Answer:

1/x+5

Step-by-step explanation:

1. Rewrite 3x as a difference

x-2/x^2 + 5x-2 - 10

2. Factor out x from the equation, then factor out -2.

x-2/ x+(x+5) - 2(x+5)

3. Factor out x+5 from the expression

x-2/ (x+5)(x-2)

4. Reduce the fraction with x-2

x-2 / (x+5) x-2

1/x+5

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2 years ago

The number of gallons of paint needed to cover a wall varies directly with the area of the wall. The Robertsons find that they u

Licemer1 [7]

Answer:

If 1/2 gallon covers 540 square feet, then 1 gallon will cover 1080 square feet.

Step-by-step explanation:

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3 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

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1 year ago

Yu and Elena each bought between 7 and 9 yards of ribbon. Yu

Mumz [18]

Given: Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon Yu bought 3 pieces of ribbon. Nailah bought 5 pieces of ribbon. Elena bought 6 pieces of ribbon.

To find: Who can buy which ribbon

Solution:

Ribbon Sizes

1 2/3 = 5/3 yard

4/5 yard

3 1/2 = 7/2 Yard

Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon

Yu bought 3 pieces of ribbon

=> 3 * 5/3 = 5

3 * 4/5 = 2.4

3 * 7/2 = 10.5

Nailah bought 5 pieces of ribbon

=> 5 * 5/3 = 8.33

5 * 4/5 = 4

5 * 7 /2 = 17.5

Elena bought 6 pieces of ribbon

=> 6 * 5/3 = 10

6 * 4/5 = 4.8

6 * 7 /2 = 21

Only value between 7 & 9 is 5 * 5/3 = 8.33

hence Nailah only can buy 1 2/3 = 5/3 yard ribbon

or there is some mistake in the data

6 0

2 years ago

Suppose a birth defect has a recessive form of inheritance. In a study population, the recessive gene (a) initially has a preval

BabaBlast [244]

Answer:

Step-by-step explanation:

a) Maternal gene = 0.25; Paternal gene = 0.25

Probability having a birth defect = 0.25 X 0.25 = 0.625

b) P(A) = 0.30; P(B) = 0.70

P(A)*P(B) = 0.30 X 0.70 = 0.021

For marriage:

25%: P(A)*P(A)*0.25 = 0.09 X 0.25 = 0.0225

65%: P(B)*P(B)*0.65 = 0.049 X 0.65 = 0.03185

10%: P(A)*0.10 = 0.03 and P(B)*0.10 = 0.07

P(A)*P(B) = 0.03 X 0.07 = 0.0021

Probability of a defect birth in the next generation = 0.0225 + 0.03185 + 0.0021 = 0.05645

c) 1. P(A)*P(A) = 0.09

2. P(B)*P(B) = 0.049

3. P(A)*P(B) = 0.021

4 0

2 years ago