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2 years ago

Which is the solution to the inaquality y+15<3

Mathematics

1 answer:

jek_recluse [69]2 years ago

5 0

Answer:

y < - 12

Step-by-step explanation:

Given

y + 15 < 3 ( subtract 15 from both sides )

y < - 12

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3 years ago

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Help!!!! for just 3 but if you know 4 also then it’s on my page and I would appreciate the help

Rus_ich [418]

A quadratic equation is set up in the form of ax² + bx +c
First set equation = to 0
0= x² - 5x - 24
Next plug into quadratic formula( -b Plus or minus the √b²-4ac) ÷ 2a
[10 plus or minus √(25² - 4×1×24)] ÷ 2
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√625- 96 = √529
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3 years ago

There are 30 blue balls, 40 red balls, and 30 white balls in a bag of balls. What is the conditional probability of choosing a b

natulia [17]

Answer: So, the probability will be

$P( \text{getting a blue ball })=\frac{30}{70}=\frac{3}{7}$

Step-by-step explanation:

Since we have given that

Number of blue balls =30

Number of red balls = 40

Number of white balls = 30

Since there are 100 balls in total ,

As we have given that there is no white ball chosen.

So, only 70 balls are left with us , from which we choose a blue ball

So, the probability will be

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3 years ago

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Show tan(???? − ????) = tan(????)−tan(????) / 1+tan(????) tan(????)<br> .

anyanavicka [17]

Answer:

See the proof below

Step-by-step explanation:

For this case we need to proof the following identity:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

We need to begin with the definition of tangent:

$tan (x) =\frac{sin(x)}{cos(x)}$

So we can replace into our formula and we got:

$tan(x-y) = \frac{sin(x-y)}{cos(x-y)}$ (1)

We have the following identities useful for this case:

$sin(a-b) = sin(a) cos(b) - sin(b) cos(a)$

$cos(a-b) = cos(a) cos(b) + sin (a) sin(b)$

If we apply the identities into our equation (1) we got:

$tan(x-y) = \frac{sin(x) cos(y) - sin(y) cos(x)}{sin(x) sin(y) + cos(x) cos(y)}$ (2)

Now we can divide the numerator and denominato from expression (2) by $\frac{1}{cos(x) cos(y)}$ and we got this:

$tan(x-y) = \frac{\frac{sin(x) cos(y)}{cos(x) cos(y)} - \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(y)}{cos(x) cos(y)}}$