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2 years ago

Mind helping a random person out with some math

Mathematics

2 answers:

Aneli [31]2 years ago

7 0

By solving it i think the answer is 7 and 8

victus00 [196]2 years ago

3 0

Answer:

may be 7 and 8

but not sure.

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Help needed asap idk anything

Licemer1 [7]

Answer:

49x +196x.... or 245x

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

7-5p+3q when p=1 and q=7

Mnenie [13.5K]

Answer:

Step-by-step explanation:

7 - 5p + 3q

p = 1, q = 7

Substitute in the values of "p" and "q", and solve;

7 - 5(1) + 3(7)

7 - 5 + 21

2 + 21

5 0

3 years ago

A right-angled triangle has shorter side lengths exactly c^2-b^2 and 2bc units respectively, where b and c are positive real num

yKpoI14uk [10]

Answer: hypotenuse = $c^{2} + b^{2}$

Step-by-step explanation: Pythagorean theorem states that square of hypotenuse (h) equals the sum of squares of each side ( $s_{1},s_{2}$ ) of the right triangle, .i.e.:

$h^{2} = s_{1}^{2} + s_{2}^{2}$

In this question:

$s_{1}$ = $c^{2}-b^{2}$

$s_{2} =$ 2bc

Substituing and taking square root to find hypotenuse:

$h=\sqrt{(c^{2}-b^{2})^{2}+(2bc)^{2}}$

Calculating:

$h=\sqrt{c^{4}+b^{4}-2b^{2}c^{2}+(4b^{2}c^{2})}$

$h=\sqrt{c^{4}+b^{4}+2b^{2}c^{2}}$

$c^{4}+b^{4}+2b^{2}c^{2}$ = $(c^{2}+b^{2})^{2}$ , then:

$h=\sqrt{(c^{2}+b^{2})^{2}}$

$h=(c^{2}+b^{2})$

Hypotenuse for the right-angled triangle is $h=(c^{2}+b^{2})$ units

3 0

3 years ago

A simple one-celled organism divides into two identical

kolezko [41]

24 single felled organisms there are to 30 minute periods in each hour which produces 2 organisms that is 12 periods and 12 multiplied by 2 is 24

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2 years ago

-6.3 as a mixed number

diamong [38]

Answer:

uit is the mixed number of decimal point

3 0

3 years ago