Using the formula , find four ordered pairs <br> F=1.8c+32

If sin x = .52 and cos x = .85, what is cot x?

Eduardwww [97]

cot x = cos x / sin x

cot x = .85 / .52 = 1.6346

6 0

2 years ago

An Ontario lottery has the following structure, the first person drawn from the names wins $15 000. Each subsequent person drawn

elixir [45]

The first four terms are : 15000, 14850, 14700, 14550
a = 15000 ; d = -150
The explicit formula is : $a_{n} = - 150n \: + 15150$

The first term of the sequence, a = 15,000

Each subsequent winner gets $150 less

Hence, the common difference, d = - $150

The first four terms :

First term, a1 = $15000

Second term, a2 = a - d = $15000 - $150 = $14850

Third term, a3 = a2 - d = $14850 - $150 = $14700

Fourth term, a4 = a3 - d = $14700 - $150 = $14550

a, for the sequence is the first term = $15000

d, for the sequence is the common $150 (difference between successive wins).

Explicit formula for the general term :

$a_{n} \: = nd \: + c$

For n = 1

$15000 = 1( - 150) + c$

$15000 = - 150 + c$

$c = 15150$

Hence, the explicit formula for the nth term is :

$a_{n} \: = - 150n \: + 15150$

Learn more : brainly.com/question/12006170

3 0

2 years ago

The ratio of oil to vinegar in a salad dressing is 8:5.How much oil must blend with 7 liters of vinegar for that recipe?

sukhopar [10]

Set it up in a proportion.

8 (oil)/5 (salad dressing) = x/7
cross multiply to get 11.2

7 0

3 years ago

Read 2 more answers

PLease help me solve this

Mnenie [13.5K]

Answer: 729

Step-by-step explanation:

$\frac{1}{3^{-2} * 3^{-4} * (-1)^{2} }$

3^-2 = 1/9

3^-4 = 1/81

(-1)^2 = 1

1/(1/9 * 1/81 * 1)

1/ 1/729 = 729

4 0

1 year ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

2 years ago

Using the formula , find four ordered pairs F=1.8c+32