PLEASE ANSWER ASAP ITS DUE!!

Find the ordered pair whose x-coordinate is 5. 4x=6y+8

sukhopar [10]

(6, 2 )

substitute x = 5 into the equation and solve for y

6y + 8 = 20 ( subtract 8 from both sides )

6y = 12 ( divide both sides by 6 )

y = $\frac{12}{6}$ = 2

the ordered pair is (5, 2 )

6 0

3 years ago

Read 2 more answers

What is 12.05 as a fraction

hram777 [196]

Answer:

241/20

Step-by-step explanation:

mmnmmmmmmnmmmmmm

4 0

3 years ago

Given that € 1 = £0.72<br> a) How much is € 410 in £?<br> b) What is the £ to € exchange rate?

Alekssandra [29.7K]

Answer:

a) 295.2

Step-by-step explanation:

a) cross multiplication method

€1=£0.72

€410=x

x=€410×£0.72

8 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

2 years ago

10. Three kinds of teas are worth $4.60 per pound, $5.75 per pound, and $6.50 per pound. They are to be

zepelin [54]

Answer:

The mass of the $4.60/lb tea that should be used in the mixture is 10 lb

The mass of the $5.75/lb tea that should be used in the mixture is 8 lb

The mass of the $6.50/lb tea that should be used in the mixture is 2 lb

Step-by-step explanation:

The parameters of the question are;

The worth of the three teas are

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of the mixture of the three teas = 20 lb

The worth of the mixture of the three teas = $5.25 per pound = $5.25/lb

The amount of the $4.60 in the mixture = The sum of the amount of the other two teas

Therefore, given that the mass of the mixture = 20 lb, we have in the mixture;

The mass of tea A + The mass of Tea B + The mass of Tea C = 20 lb

The mass of tea A = The mass of Tea B + The mass of Tea C

Therefore;

The mass of tea A + The mass of tea A = 20 lb

2 × The mass of tea A in the mixture = 20 lb

The mass of tea A in the mixture = 20 lb/2 = 10 lb

The mass of tea A in the mixture = 10 lb

The mass of Tea B + The mass of Tea C = The mass of tea A = 10 lb

The mass of Tea B + The mass of Tea C = 10 lb

The mass of Tea B = 10 lb - The mass of Tea C

Where the mass of Tea C in the mixture = x, we have;

The mass of Tea B in the mixture = 10 lb - x

The cost of the 10 lb of tea A = 10 × $4.60 = $46.0

The worth of the tea mixture = 20 × $5.25 = $105

The worth of the remaining 10 lb of the mixture comprising of tea A and tea B is given as follows;

The worth of Tea B + The worth of Tea C in the mixture = $105.00 - $46.00 = $59.00

Therefore, we have;

x lb × $6.50/lb + (10 - x) lb × $5.75/lb = $59.00

x × $6.50 - x × $5.75 + $57.50 = $59.00

x × $0.75 = $59.00 - $57.50 = $1.50

x = $1.50/$0.75 = 2 lb

∴ The mass of Tea C in the mixture = 2 lb

The mass of Tea B in the mixture = 10 lb - x = 10 lb - 2 lb = 8 lb

The mass of Tea B in the mixture = 8 lb

Therefore, since we have;

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of tea A in the mixture = 10 lb

The mass of tea B in the mixture = 8 lb

The mass of tea C in the mixture = 2 lb, we find;

The mass of the $4.60/lb tea that should be used in the mixture = 10 lb

The mass of the $5.75/lb tea that should be used in the mixture = 8 lb

The mass of the $6.50/lb tea that should be used in the mixture = 2 lb.

6 0

3 years ago