What are all these . Please help I will mark Brainly

Mathematics

1 answer:

disa [49]3 years ago

3 0

Answer:

for a the answer is 5 for b also 5 (the solution is also the same but the = sign will be >) and for c is true

Step-by-step explanation:

for a so its 2x=10 you divide it like this

<h3><u>2x</u>=<u>10</u></h3><h3>2 2 so the answer will be</h3><h3>x=5 so you mark 5</h3>

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Can 12 cm 12 cm and 28 cm form a triangle

erastova [34]

Using the Theorem of Pythagoras (a² + b² = c²)

Put in the values:

12² + 12² = 28²
Is it correct? (solve)

Nope! Because 12² + 12² = 288, and the square root of 288 is about 17, not 28.

The answer is no, 12cm, 12cm, 28cm cannot form a triangle.

Hopefully this helps! If you have any more questions or don't understand, feel free to DM me, and I'll get back to you ASAP! :)

3 0

3 years ago

1)How many ways can the letters in the word BOOKKEEPER be arranged? (Must show set-up )

sergeinik [125]

Question 1

There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are $\frac{10!}{(10-6)!6!}$ ways of arranging the letters, which equal to 210 ways

Question 2

There are seven swimmers in total.
There are $\frac{7!}{(7-1)!1!}$ ways of choosing the first winner, which is 7 ways
There are $\frac{6!}{(6-1)!1!}$ ways of choosing the second winner, which is 6 ways
There are $\frac{5!}{(5-1)!1!}$ ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner

Question 3

The probability of eating an orange and a red candy is $\frac{15}{31}$ × $\frac{9}{30}$ , which equals to $\frac{9}{62}$

The probability of eating two green candies is $\frac{7}{31}$ × $\frac{6}{30}$ which equals to $\frac{7}{155}$

3 0

3 years ago

#82 will give brainliest to best answer!

Fofino [41]

Answer:

$\frac{6}{k-6}$

Step-by-step explanation:

First, we can factor all of the following equations to turn that weird, huge looking thing into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ ÷ $\frac{(k-6)^2}{k(k-6)}$ × $\frac{6(k-10)}{k(k + 6)}$ . We know that division is simply multiplication by the reciprocal, so that whole equation will turn into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ × $\frac{k(k-6)}{(k-6)^2}$ × $\frac{6(k-10)}{k(k+6)}$ . Now we can cancel out some values if they are both in the numerator and denominator, which will turn that still huge looking thing into $\frac{6}{k-6}$ which is our final answer, as it cannot be simplified further.