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3 years ago

Do the ordered pairs (2, 20), (5, 15), (12, 19), and (14, 13) represent a function?

Mathematics

2 answers:

Lilit [14]3 years ago

6 0

Yes it does because the c values don’t repeat

Natalka [10]3 years ago

3 0

Answer:

yes

Step-by-step explanation:

You might be interested in

A cone has a height of 1 meter and a diameter of 2 meters. What is it’s volume

Ber [7]

Cone volume formula: V = πr²h/3

r = radius

h = height

The radius is half the diameter, so, we can divide.

2 / 2 = 1

Now, solve with the given values.

V = π(1)²(1)/3

V = π(1)(1/3)

V = 3.14(1/3)

V ≈ 1.05

Therefore, the volume is roughly 1.05m^3

Best of Luck!

6 0

4 years ago

Will give out 25 points for right answer!!! (Both parts of answer 1)

patriot [66]

Answer:

1.x=41

x=9

Step-by-step explanation:

1.x-7=34

x=41

3x-7=20

3x=27

x=9

3 0

1 year ago

Find the percentage of vacationers from<br> question 10 who spent between $1500<br> and $2000.

mars1129 [50]

Answer:

The percentage change of vacationers is 33.3 %

Step-by-step explanation:

Given as :

The old value of the vacationers = $ 1500

The new value of the vacationers = $ 2000

Let the percentage variation = x %

Or, x % increase = $\dfrac{\tyextrm new value - \textrm old value}{\textrm old value}$ × 100

Or, x % increase = $\dfrac{\tyextrm $ 2000 - \textrm $ 1500}{\textrm $ 1500}$ × 100

Or, x % increase = $\frac{500}{1500}$ × 100

Or, x % increase = $\frac{1}{3}$ × 100

Or, x % increase = $\frac{100}{3}$

∴ x = 33.3 %

So, The percentage increase change is 33.3 %

Hence The percentage change of vacationers is 33.3 % Answer

3 0

3 years ago

I need help... on my homework it says to write each expression in radical form, or write each radical and exponential form . For

pantera1 [17]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt{13}\implies \sqrt[2]{13^1}\implies 13^{\frac{1}{2}}$

5 0

3 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

4 0

4 years ago