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3 years ago

15

Veronica's fall football league has a phone tree in case games or practices are postponed. The coach calls 3 players. Then, each

of those players calls 3 players, and so on. How many players will be notified during the third round of calls

2 answers:

zepelin [54]3 years ago

7 0

Answer:

39

Step-by-step explanation:

the first three call 3, making it 3*3, then each of them will get 3 more, making it 3+(3*3)+(3*3*3),3+9+27, 39

Leviafan [203]3 years ago

4 0

Answer: 9

Step-by-step explanation:

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(I do brainliest.) slope intercept form... [show ya work plz]

eduard

Answer:

y = 3x+2

Step-by-step explanation:

(3,7) (6,9) Y2=9 Y1=7 X2=6 X1=3

M = 9-7/6/3

2/3

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3 years ago

The box-and-whisker plot summarizes student scores on a physical fitness test. Students receive an award for scores at or above

Lena [83]

Answer:

74

Step-by-step explanation:

To receive an award, student must obtain a score at or above the upper quartile, from.tye boxplot, the upper quartile is depicted by the endpoint of the box. Hence, the upper quartile of the distribution is 74. Hence only stude Ts who score 74 and above will receive an award Hence, the minimum score required is 74.

4 0

3 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

4 0

3 years ago

diamong [38]

Answer:

Cos A=5/13

we have

Cos² A= $1-Sin ²A$

25/169=1-Sin²A

sin²A=1-25/169

sin²A=144/169

Sin A= $\sqrt{144/169}=12/13$

again

Tan B=4/3

P/b=4/3

p=4

b=3

h= $\sqrt{3²+4²}=5$

Now

Sin B=p/h=4/5

in IV quadrant sin angle is negative so

Sin B=-4/5

CosB=b/h=3/5

Now

<u>S</u><u>i</u><u>n</u><u>(</u><u>A</u><u>+</u><u>B</u><u>)</u><u>:</u><u>s</u><u>i</u><u>n</u><u>A</u><u>c</u><u>o</u><u>s</u><u>B</u><u>+</u><u>C</u><u>o</u><u>s</u><u>A</u><u>s</u><u>i</u><u>n</u><u>B</u>

<u>n</u><u>o</u><u>w</u><u> </u>

<u>substitute</u><u> </u><u>value</u>

<u>Sin(A+B):</u>12/13*3/5+5/13*(-4/5)=36/65-4/13

<u>=</u><u>1</u><u>6</u><u>/</u><u>6</u><u>5</u><u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u>

6 0

3 years ago

A child is 2 -1/2 feet tall. The child’s mother is twice as tall as the child. How tall is the child’s mother

Len [333]

Answer:

5 feet

Step-by-step explanation:

"Twice as tall" means "2 times as tall".

2 × (2 1/2 ft) = (2 × 2 ft) +(2 × (1/2 ft)) = 4 ft + 1 ft = 5 ft

The child's mother is 5 feet tall.

7 0

3 years ago

Read 2 more answers