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S_A_V [24]
3 years ago
14

Need help on both of these for geometry

Mathematics
1 answer:
artcher [175]3 years ago
4 0

Answer:

Step-by-step explanation:

have the triangles written in the same order to se the corresponding angles

C      A .   T

⇵ .  ⇵ .   ⇵

D      O .   G

∠D ≅∠C

∠O ≅∠A

GD ≅TC

10.

you have an isosceles triangle with vertex angle 48 so the base angle is

(180-48)/2 = 66

32+ x = 66 because is an exterior angle

x= 66-32 = 34

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Please help if you help you are swag!!
Ostrovityanka [42]

Answer:

I THINK it's A but i might be wrong, good luck tho

Step-by-step explanation:

3 0
3 years ago
You have 22 coins, dimes and nickels. If the number of dimes and nickels were reversed, you would have $0.40 less than you actua
fenix001 [56]
The coins differ in value by 5¢, so swapping the numbers of them will change the value by 5¢ for each unit difference in the numbers of coins. Since
40¢ = 8 * 5¢
there must be 8 more dimes than nickels.

There are (22 +8)/2 = 15 dimes and 7 nickels.

_____
You could write some equations for this problem. Let n, d represent numbers of nickels and dimes.
.. n +d = 22
.. 10d +5n - (10n +5d) = 40 . . . . . . . cents
.. 5(d -n) = 40 . . . . . . . . . . . . . . . . . . reversing the coin count changes the total by 5 cents for each unit of difference (d -n), as stated above.
.. d -n = 8 . . . . . . . . . . . . . . . . . . . . . . divide the preceding equation by 5
Adding this last equation to the first gives
.. 2d = 22 +8
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.. n = 22 -15 = 7
There are 15 dimes and 7 nickels.
4 0
4 years ago
On a multiple choice test, if you randomly guessed on three questions, then what is the probability you got at least one of them
snow_lady [41]

Answer:

Number of Questions =3

Probability of giving a correct answer

                              =\frac{1}{3}

Probability of giving two correct answers

                       =\frac{2}{3}

Probability of giving all correct answers

            =\frac{3}{3}\\\\=1

Probability that at least one of them is correct

         =_{1}^{3}\textrm{C}\\\\=\frac{3!}{(3-1)! \times1!}\\\\=3 \text{ways}\\\\=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \\\\=\frac{1}{27}

Probability that two of them is correct

         =_{2}^{3}\textrm{C}\\\\=\frac{3!}{(3-2)! \times2!}\\\\=3 \text{ways}\\\\=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\\\=\frac{8}{27}

Probability that all of them is correct

         =_{3}^{3}\textrm{C}\\\\=\frac{3!}{(3-3)! \times3!}\\\\=1 \text{way}\\\\=1

So, Required probability

         =\frac{1}{27} \times \frac{8}{27} \times 1\\\\=\frac{8}{729}

6 0
3 years ago
I WILL GIVE BRAINLIEST to who answers this question correctly. Please help I need it done tonight!
podryga [215]
Four quarters are .066 thick
20 nickels are 1.48 thick
<span>So your answer is 1.21 inches thicker or about 1 1/4 thicker</span>
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Answer:

not enough to figure out :/

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