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2 years ago

Calculate the unit rate:write your answer as a decimal 1/2 mile in ever 1/3 hour

Mathematics

1 answer:

Alexandra [31]2 years ago

7 0

Answer:

$unit \: rate = \frac{distance}{time}$

unit rate is similar to speed / velocity:

$unit \: rate = \frac{ (\frac{1}{2} )}{ (\frac{1}{3})} \\ \\ = \frac{1}{2} \div \frac{1}{3} \\ \\ = \frac{1}{2} \times \frac{3}{1} \\ \\ = > { \boxed{ \boxed{unit \: rate = \frac{3}{2} \: miles \: per \: hour}}}$

[unit rate must have units including per <time> ]

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Hey what's <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B169%7D%20%20%5Ctimes%202" id="TexFormula1" title=" \sqrt{169} \ti

WITCHER [35]

Answer:

the square root of 169 is 13 then ×2

I hope I am correct mate

enjoy your day

#Captainpower

4 0

2 years ago

Read 2 more answers

The price of a company's share dropped by 3.50% by the end of the first year, down to $44.25. During the second year the price o

omeli [17]

Answer:

a. $45.86

b. $43.87

c. Percentage change = -4.40%)

Step-by-step explanation:

a. Let the price of share at the beginning of first year be represented with X

Now, Price of company's share dropped by 3.50% at the end of first year.

So, Price of share at the end of first year = x - 3.5% of x

= x - 0.035x

= 1x - 0.035x

= 0.965x

But it is equal to $44.25

=> 0.965x = 44.25

x = 44.25 / 0.965

x = 45.8549223

x = $45.86

b. During second year price of share decreased by $1.99. Therefore, Price of share at the end of second year = $45.86 - $1.99 = $43.87

c. Percentage change in the price of shares over two years = {(45.86 - 43.87)/45.86} *100

= (1.99/45.86)*100

= 0.04339294 * 100

= 4.40%

Now, as price of shares has dropped, the percentage change will be negative. (Δ% = -4.40%)

5 0

2 years ago

What is the 7th term in the sequence of square numbers?

12345 [234]

Answer:

49.

Step-by-step explanation:

To solve, you can simply list the square numbers in order...

1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2.

1, 4, 9, 16, 25, 36, 49.

Hope this helps!

5 0

3 years ago

Read 2 more answers

Find the area of the region enclosed by the graphs of the functions

Vaselesa [24]

Answer:

$\displaystyle A = \frac{8}{21}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Terms/Coefficients
Functions
Function Notation
Graphing
Solving systems of equations

Calculus

Area - Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

*Note:

Remember that for the Area of a Region, it is top function minus bottom function.

Step 1: Define

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

Step 2: Identify Bounds of Integration

Find where the functions intersect (x-values) to determine the bounds of integration.

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

Step 3: Find Area of Region

Integration

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx$
[Area] Rewrite [Integration Property - Subtraction]: $\displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx$
[Area] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = \frac{2}{3} - \frac{2}{7}$
[Area] Subtract: $\displaystyle A = \frac{8}{21}$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

6 0

3 years ago

After a rotation of 90° about the origin, the coordinates of the vertices of the image of a triangle are A'(6, 3), B'(-2, 1),

Rasek [7]

Answer:

A(-3, 6), B(-1, -2), C(-7, 1)

Step-by-step explanation:

To the pre-image after a 270°-counterclockwise rotation [90°-clockwise rotation], just reverse it by doing a 270°-clockwise rotation [90°-counterclockwise rotation]:

Extended Rotation Rules

270°-clockwise rotation [90°-counterclockwise rotation] >> (x, y) → (-y, x)
270°-counterclockwise rotation [90°-clockwise rotation] >> (x, y) → (y, -x)
180°-rotation >> (x, y) → (-x, -y)

So, perform your rotation:

270°-clockwise rotation [90°-counterclockwise rotation] → C'[1, 7] was originally at C[-7, 1]

→ B'[-2, 1] was originally at B[-1, -2]

→ A'[6, 3] was originally at A[-3, 6]

I am joyous to assist you anytime.

3 0

3 years ago

Read 2 more answers

Calculate the unit rate:write your answer as a decimal 1/2 mile in ever 1/3 hour​

Calculate the unit rate:write your answer as a decimal 1/2 mile in ever 1/3 hour