Please help I will mark...BRAINLIEST

Mathematics

2 answers:

frozen [14]3 years ago

6 0

Answer:

I would but...

Step-by-step explanation:

where the question at

sergey [27]3 years ago

6 0

Answer:

where is the question what do you need help with

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The fraction 7/9 produces a repeating decimal 0.7 A. true B. false

Elis [28]

Answer:

false for 0.7 it would have to be 7/10

Step-by-step explanation:

7 0

3 years ago

Find the hypotenuse of each isosceles right triangle when the legs are of the given measure. 6 sqrt 2

Gemiola [76]

Isosceles right triangles have two equal sides (a and b) that are not the hypotenuse (c). And when two sides are equal, so are their opposite angles. There are only 180° degrees in any triangles, thus the right angle = 90°, so 90 left for the two equal, means that 2x=90,
x = 45°.

There are several ways to go about solving a triangle like this. The best and easiest is simply to memorize that the hypotenuse is exactly root2 times the other sides. Or, each isosceles side is the hypotenuse (c) ÷ root2
$a = b = c \div \sqrt{2} \\ c = a\sqrt{2} \\ c = 6 \sqrt{2} \times \sqrt{2} = 6 \times 2 = 12$
Another way to do it is the longer proof of Pythagorean Theorem:
${c}^{2} = {a}^{2} + {b}^{2}... \: \: c = \sqrt{({a}^{2} + {b}^{2})} \\$
$c= \sqrt{({6 \sqrt{2}) }^{2} + ({6 \sqrt{2})}^{2}} \\ = \sqrt{(2 \times{(6 \sqrt{2} )}^{2} )} = \sqrt{2(36 \times 2)} \\ c = \sqrt{144} = 12$

7 0

3 years ago

Chung is practicing archery so far he has fired 5 arrows at a target and hit the center 2 times at this rate if he fires 100 arr

bagirrra123 [75]

40 because if you multiply 2 by 100 it is 200 then divide that by 5 and you get 40

4 0

3 years ago

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HELP FAST SO CONFUSED Find the area of the shaded sector. Leave your answer in terms of π.

ddd [48]

Total area: pi*9^2 = 81pi

Shaded area = Total area / 3, because 120/360 = 1/3

So Shaded area = 81pi/3 = 27pi

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3 years ago

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Find a normal subgroup of v which is not normal in a4.

Tpy6a [65]

he elements of the Klein 44-group sitting inside A4A4 are precisely the identity, and all elements of A4A4of the form (ij)(kℓ)(ij)(kℓ) (the product of two disjoint transpositions).

Since conjugation in SnSn (and therefore in AnAn) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.

6 0

3 years ago