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2 years ago

8

Jeffrey ate out for lunch his meal cost 10.50 he would like to leave an 18% tip. how much does the lunch cost in total?

1 answer:

Mila [183]2 years ago

5 0

Answer:

12.39

Step-by-step explanation:

18% 10.50 = 1.89

10.50+1.89 = 12.39

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Find the volume please and Katie, why are you deleting my questions?

Rina8888 [55]

Volume of the prism = 75 cubic feet

Solution:

The image given is a rectangular prism (or) a cuboid.

Area of base = l × b = 25 square feet

Height of the prism = 3 feet

Volume of the prism = l × b × h

= (l × b) × h

= 25 × 3

= 75 cubic feet

Volume of the prism = 75 cubic feet

Hence, the volume of the prism is 75 cubic feet.

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3 years ago

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Marty bought 4 new books today. he plans to read 2 of these while on vacation. how many combinations of 2 books from the group o

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6 :)

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3 years ago

A tank can hold 300 cm^3 of water. How much would Tim have to pay If he has to fill 12 such tanks with water at 25 cents per cm^

katrin [286]

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3 years ago

Which ordered pair can be plotted together with these four points, so that the resulting graph still represents a function?

MrRissso [65]

Answer:

B.(-1,2)

Step-by-step explanation:

In a function, there can not be two different values of y corresponding to the same value of x.

See the graph attached.

Here, the points on the graph are (1,2), (2,-3), (-2,-2) and (-3,1).

If we consider point (-2,2) then there will be two points corresponding to the same x value i.e. (-2,-2) and (-2,2).

Similarly, if we consider the point (2,-2) or (2.-1) then also there becomes more than one values of y for a single value of x i.e. x = 2.

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2 years ago

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

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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3 years ago