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3 years ago

I need help with this algebra question

Mathematics

2 answers:

ioda3 years ago

7 0

Answer:

4.58258

Step-by-step explanation:

Pythagorean Theorem:

a² + b² = c²

Input:

2² + x² = 5²

Remember, c² is always the side opposite of the right angle...

4 + x² = 25

21 = x²

x = 4.58258

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

dybincka [34]3 years ago

3 0

Answer:

$\sqrt{21}$

Step-by-step explanation:

let the missing side be x , then

Using Pythagoras' identity in the right triangle.

x² + 2² = 5²

x² + 4 = 25 ( subtract 4 from both sides )

x² = 21 ( take the square root of both sides )

x = $\sqrt{21}$ ≈ 4.6 ( to 1 dec. place )

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Work out the size of angle x Next to 164 and 90

Zielflug [23.3K]

Answer:

x=106 degrees

Step-by-step explanation:

I'm going to assume that this is a circle since you didn't give it to me

So you want to add 164 and 90 together and subtract the sum by 360

360-(164+90)=106

So you want to use PEMDAS (order of operations) to solve this

so do what's in the perenthesis first so 164+90=254
Then you want to subtract 254 from 360 so 360-254=106

x=106 degrees

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3 years ago

What is 12 divided by 1/6

TEA [102]

The answer would be, 72.

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3 years ago

A coin is tossed 10 times resulting in 7 heads and 3 tails. The same coin is tossed 1000 times resulting in 510 heads and 490 ta

KiRa [710]

The answer is C. 50%.

The theoretical probability has nothing to do with the experiments. So, we will forget results of the experiment and think about theoretical probability. A coin has two sides - head and tail. The probability to get head is 1/2 = 0.5 = 50%. This is because if you toss the coin and you get head, head is one probability of two probability in total (head and tail). The same situation is with tail. Tail is .one probability of two probability in total (head and tail).

8 0

3 years ago

If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (

AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

$P=\frac{3}{36}$

$P=\frac{1}{12}$

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

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3 years ago

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coldgirl [10]

Answer:

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Step-by-step explanation:

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