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attashe74 [19]
2 years ago
6

Who wants 100 points you also get brainliest if you answer this question... 6+9=?

Mathematics
1 answer:
FrozenT [24]2 years ago
6 0

Answer:

6+9=15

Step-by-step explanation:

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What is the axis of symmetry of the function f(x)= -(x+9)(x-21)
Vinil7 [7]

Answer:

The axis of symmetry is x = 6

Step-by-step explanation:

To find this, first find the two x-intercept values and then take the average. This will always be the line of symmetry.

x + 9 = 0

x = -9

x - 21 = 0

x = 21

Now take the average of these two numbers.

(-9 + 21)/2

12/2

6

6 0
3 years ago
Can someone help me find x please
a_sh-v [17]

Answer:

38.7°

Step-by-step explanation:

We can solve the triangle using the trigonometrical ratios which may be expressed in the form SOA CAH TOA Where

SOA stands for

Sin Ф = opposite side/hypotenuses side

CAH stands for

Cosine Ф = adjacent side/hypotenuses side

TOA stands for

Tangent Ф = opposite side/adjacent side

The hypotenuse is the side facing the right angle while the opposite is the side facing the given angle.

Considering the triangle with respect to angle x, 8cm is the opposite side, 10cm is the adjacent side

hence

Tan x = 8/10

Tan x = 0.8

x = tan -1 0.8

= 38.7°

4 0
3 years ago
8q + 6 = 4q - 14<br> Please help
ICE Princess25 [194]

Answer:

q=-5

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
17 cm
Agata [3.3K]
D ststjdkgdgk dhdgkdlydky
6 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
2 years ago
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