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zimovet [89]
2 years ago
14

What number makes the equation true? [???] ÷ 8 = 4​

Mathematics
2 answers:
DochEvi [55]2 years ago
5 0

Answer: 32

Step-by-step explanation:

If you divide 32 by 8 then you get 4

Elina [12.6K]2 years ago
3 0

Answer:

32

Step-by-step explanation:

math

................

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I would but my teacher has a ad blocker on and I cant access the attachment :P


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3 years ago
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A necklace regularly sells for $18.00. the store advertises a 15% discount. what is the sale price of the necklace in dollars
SSSSS [86.1K]
I can solve with 2 methods:

I. Because the discount is 15 % of 18 $ , the price will be (100-15=85) 85 % of 18 $
18*85/100= 15,3 $ ( the sale price)

II. The discount is 15 % of 18$
15*18/100= 2,7$  ( the discount)
then I decrease it from the regularly price
18$-2,7$=15,3 $ (the sale price)

Personally I believe the first is an easier method.
I hope you understand and you can apply this in every similary problem.



6 0
3 years ago
Shiela is 1.7m tall. Her son is 109cm tall. How many meters taller is Shiela than her son?
Tju [1.3M]
Shiela is 0.61 meters taller than her son.
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3 years ago
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Based on the Pythagorean Theorem, which of
aleksley [76]

Answer:

G is not TRUE.

Step-by-step explanation:

using the law A+B = B+A

PYTHAGOREAN THEOREM

A²+B² =C² is equal to B² +A² = C²

So for A²,

B² - c² = A. remember if a positive number move from the left to the right over an equal sign it becomes negative and vice versa

B²

C² - A²= B²

6 0
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
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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

7 0
3 years ago
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