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2 years ago

I need help bad I am about to fail my senior year if I can’t get all my work done.

Mathematics

1 answer:

Nikolay [14]2 years ago

3 0

a=9/14

b=32/8

c=25/13

That is the answer for these questions.

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2 Solve for a. 5 a = √ [?] a = Pythagorean Theorem: a² + b2 = c²

yuradex [85]

Answer:

$a = \sqrt{21}$

Step-by-step explanation:

Pythagorean theorem
$\sqrt{a ^{2} + b ^{2} } = c \\ \sqrt{c ^{2} - b ^{2} } = a$
$\sqrt{5 ^{2} - 2 ^{2} } = \sqrt{21}$

4 0

2 years ago

Read 2 more answers

Gina has a jar of marbles that contains 7 blue marbles, 5 red marbles, and 8 green marbles. She randomly selects a marble, recor

BigorU [14]

The probability that Gina randomly selected two red marbles is 1/19

Explanation:

Total number of marbles = 7 + 5 + 8

= 20

The probability of getting two red marbles in the fraction form is given as:

P(first red marble) = number of red marbles / total number of marbles

P(first red marble) = $\frac{5}{20} = \frac{1}{4}$

P(second red marble) = number of red marbles after 1 white marble is removed / total number of marbles after 1 red marble is removed.

P(first red marble) = $\frac{4}{19}$

P(two red marbles) = P(first) X P(second)

= $\frac{1}{4} X \frac{4}{19}$

= $\frac{1}{19}$

Therefore, the probability that Gina randomly selected two red marbles is 1/19

6 0

3 years ago

Solve the following system by the comparison method.

agasfer [191]

Answer:

2p + q = 1

9p + 3q + 3 = 0

q = 1 - 2p

replace q = 1 - 2p into 9p + 3q + 3 = 0

9p + 3(1 - 2p) + 3 = 0

9p + 3 - 6p + 3 = 0

3p + 6 =0

3p = -6

p = -2

q = 1 - 2p

q = 1 -2(-2)

q = 1 + 4

q = 5

(-2 , 5)

there for the answer would be

{(-2, 5)}

Hopefully this was helpful <3 :3

5 0

3 years ago

HELP PLS the dimensions of an triangle are shown

solong [7]

Answer:

Answer is B ( 314.9 m2)

Step-by-step explanation:

A= H(B)/2

A= 25.6(24.6)/2

A= 314.88

6 0

3 years ago

Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

3 years ago