300/200 as a terminating decimal.

Mathematics

2 answers:

Mumz [18]2 years ago

4 0

Answer:

1.5

Step-by-step explanation:

Tomtit [17]2 years ago

3 0

Answer:

1 100/200

Step-by-step explanation:

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Which is mood supported by the passage ?

zepelin [54]

The correct answer is A. Longing I know because I just finished the quiz.

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2 years ago

What is the slope of the line that passes through the points(7,-9) and (15,-29)?

Alex Ar [27]

(Y-Y1)/(X-X1)
(-9-(-29))/(7-15)
20/-8
-2 1/2
The slope is -2 1/2

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2 years ago

Minny has two coupons, but can use one of them. The first one is for $5 off an item; the second coupon is for 20% off an item. T

joja [24]

Let's first solve for the $5 off coupon,

$28 - 5 = 23$

The final total is $23.00 with $5 off.

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Now for the final price of the 20% off coupon,

$20\% = 0.2$

0.2 or 0.20 is decimal form of 20%. Easier to multiply than in percent form!

$0.2 \times 28 = 5.6$

$5.6 or $5.60 is 20% of 28, we're not done yet!

Lastly, subtract 20% off of 28 (5.6) from 28,

$28 - 5.6 = 22.4$
$22.4 or $22.40 is the final after 20% off!

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Q. Which is less after the discounts??

$23.00 or $22.40

A. $22.40 of course which was the 20% off.

Part A : Which coupon should she use ?

Answer : Well, to get a lower price, less to pay, the 20% off coupon is better!

Part B: How much money will she save by using the better coupon?

Answer : $5.6 / $5.60 is the amount she'll save!

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I hope that helps you out!!

Any more questions, please just ask me!!

~ Zoey

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2 years ago

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$