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3 years ago

Darnell took a total of 14 pages of notes during 7 hours of class. After attending 8 hours of

Mathematics

1 answer:

SVEN [57.7K]3 years ago

4 0

Answer:

16 pages

Step-by-step explanation:

1 hour= 2 pages written of notes

so 8 hours of notes = 8x2=16 pages

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Check thepic plzz first answer brainliest

aleksandr82 [10.1K]

Answer:

Total Surface Area: A = 88 square cm

Step-by-step explanation:

Volume V = 48 cubic cm

So this cuboidal box is actually a rectangular prism.

V = L * W* H

L = length, W = breadth, H= height

With L: W : H = 3: 2: 1

L:W = 3: 2

W: H = 2: 1

L/W = 3/2, L = (3/2)*W

W/H = 2/1 , W = 2*H

L = (3/2)*(2H) = 3H

So : V = L * W * H = (3H) * (2H) * H = 6 * H^3

48 = 6 * H^3

48/6 = 8 = H^3

H = cube-root(8) = 2 cm

so ...

W = 2*H = 2*2 = 4 cm

L = (3/2)*W = (3/2)* 4 = 6 cm

Total Surface Area: A = 2*LW + 2*LH + 2*WH

A = 2*(6 * 4) + 2*(6 * 2) + 2*(4 * 2)

A = 2*24 + 2*12 + 2*8

A = 48 + 24 + 16

A = 48 + 40

A = 88 square cm

4 0

4 years ago

Simplify the expression.

Amanda [17]

Answer:

sorry I don't understand that I will answer you later sorry

6 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago

Which is the ratio of the number of months that begin with the letter J to the total number of months in a year? (2 points) 12 t

jeka57 [31]

Answer:

3-12

Step-by-step explanation:

January

June

July

That is 3 months out of 12 so...3 to 12

8 0

4 years ago

One lap around the schools track is 1/4 mile. Tyler ran two times around the track. Then he ran 5/6 mile home. How far did tyler

Mrac [35]

8/6 or 1 2/6 or 1 1/4

Explanation:

1/4 + 1/4 = 2/4

2/4 = 3/6

5/6 + 3/6 = 8/6

8/6 simplified is 1 2/6 or 1 1/4

6 0

3 years ago