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1 year ago

I neeeeeed help pls 6th grade easy

Mathematics

2 answers:

Julli [10]1 year ago

6 0

The information that is not needed to determine the total bill after tax is is the tax was 7%.

The total bill is $11.

<h3>What is the total bill?</h3>

The total bill paid is the sum of the cost of the meal, tax and the tip paid. The tip and tax would increase in the total amount that is paid by the customers.

Tax is a compulsory amount that is levied by the government on certain goods and services.

Percentage is the fraction of an amount expressed as a number out of hundred. To convert a number from percentage to a number, divide the percentage by 100.

Tax paid = 10% x total bill

$1.10 = 0.1 x b

1.10 = 0.1b

b = 1.10 / 0.1

b = $11

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gavmur [86]1 year ago

3 0

a. The information that is needed to solve the problem is D. All of this information is needed to solve the problem.

b. The total bill is $117.

<h3>What is an expression?</h3>

It should be noted that an expression shows the relationship between the numbers or the variables

The total bill will be:

Tip = $1.10

Tax = 7%

Tip = 10% of cost:

$10 = 10% × c

0.1c = $10

c = $10/0.1 = $100

Cost = $100

The total bill. = Cost + Tip + Tax

= $100 + $10 + (7% × $100)

= $100 + $10 + $7

= $117

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Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so