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LUCKY_DIMON [66]

3 years ago

7

Noah starts with 0 and then adds the number 5 four times. Diego starts with 1 and then multiplies by the number 5 four times. Fo

r each expression, decide whether it is equal to Noah’s result, Diego’s result, or neither.
4 ⋅ 5

4 + 5

4^5

5^4

1 answer:

Fittoniya [83]3 years ago

4 0

4 • 5 is Noah and 5^4 is Diego

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If 5 potatoes together have a mass of 1 kg and 8 pears together have a mass of 1,200 g which one has a greater mass, a potato or

klasskru [66]

8 ''Pears have a greater mass'' because 1000g is equal to 1kg and so the pears are 1,200g it would be 1.2k > 1k... hope it helps you :)

6 0

3 years ago

A block of ice has a square top and bottom and rectangular sides. At a certain point in

Lilit [14]

Answer:

$\frac{dV}{dt} = 360 \ cm^3/h$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

<u>Geometry</u>

Volume of a Rectangular Prism: V = lwh

<u>Calculus</u>

Derivatives

Derivative Notation

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

<u /> $l = 30 \ cm\\w = l\\\frac{dl}{dt} = 2 \ cm/h\\h = 20 \ cm\\\frac{dh}{dt} = 3 \ cm/h$ <u />

<u />

<u>Step 2: Differentiate</u>

Rewrite [VRP]: $V = l^2h$
Differentiate [Basic Power Rule]: $\frac{dV}{dt} = 2l\frac{dl}{dt} \frac{dh}{dt}$

<u>Step 3: Solve for Rate</u>

Substitute: $\frac{dV}{dt} = 2(30 \ cm)(2 \ cm/h)(3 \ cm/h)$
Multiply: $\frac{dV}{dt} = 360 \ cm^3/h$

Here this tells us that our volume is decreasing (ice melting) at a rate of 360 cm³ per hour.

3 0

3 years ago

loris [4]

Answer:

12 4 3232qwe

Step-by-step explanation:

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

identify the coordinates of four points on the line with each given slope and y-intercept. slope= 1.5, y-intercept= -3

Whitepunk [10]

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

We have the slope = 1.5 and y-intercept b = -3. Substitute:

$y=1.5x-3$

Choose any value of x and calculate y:

for x = -2

$y=1.5(-2)-3=-3-3=-6\to(-2,\ -6)$

for x = -1

$y=1.5(-1)-3=-1.5-3=-4.5\to(-1,\ -4.5)$

for x = 0

$y=1.5(0)-3=0-3=-3\to(0,\ -3)$

for x = 1

$y=1.5(1)-3=1.5-3=-1.5\to(1,\ -1.5)$

for x=2

$y=1.5(2)-3=3-3=0\to(2,\ 0)$

<h3>Answer: (-2, -6); (-1, -4.5); (0, -3); (1, -1.5); (2, 0)</h3>

7 0

4 years ago