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3 years ago

{-14, -8, -3, 2, 12, 24, 35}

Mathematics

1 answer:

crimeas [40]3 years ago

3 0

It would be the b answer hope it works out for you

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What is the solution for tgx >=0?

ddd [48]

well, what variable are you solving for? like if you're solving for x, it would be:

tg is greater than or equal to 0x

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3 years ago

Declan's friend Luka claims that he can read minds. To test Luka's abilities, Declan draws 5 cards without replacement from a st

gogolik [260]

Answer:

Step-by-step explanation:

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3 years ago

The coordinates of the points A and B are (0,-2) and (3,0), what is the y intercept of the line AB

lilavasa [31]

Answer:

The y-intercept of points A and B is -2.

Step-by-step explanation:

Point A is the y-intercept. You can tell by looking at the x and seeing that it is 0. This means the point is the y-intercept.

Hope this helps!

-Luna

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3 years ago

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Why do some quadratic graphs move left or right and others do not? Is there a correlation between this and them moving up and do

Vikentia [17]

Answer:

(x + 3)^2 is left 3 because it becames 0 at -3

(x - 3)^2 is right 3 because it becames 0 at 3

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x^2 - 3 is down 3 because your just subtracting 3 units down

Step-by-step explanation:

8 0

3 years ago

50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle

aalyn [17]

Answer:

$EF=4\sqrt{3}$

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

$A_{\text{rect}}=8(6)=48\text{ cm}^2$

The area of the triangle is 1/4 of this. Therefore:

$\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2$

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

$\displaystyle 12=\frac{1}{2}(DE)(DF)$

Since DE = DF:

$24=DF^2$

Thus:

$DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE$

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

$(DE)^2+(DF)^2=(EF)^2$

Therefore:

$(2\sqrt6)^2+(2\sqrt6)^2=EF^2$

Square:

$24+24=EF^2$

Add:

$EF^2=48$

And finally, we can take the square root of both sides:

$EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}$

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3 years ago

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