All categories

2 years ago

Help it’s my last question

Mathematics

1 answer:

Digiron [165]2 years ago

6 0

Answer:

alr so basically what you want to do is multiply the numbers to get the answer

Step-by-step explanation:

You might be interested in

Sal has already finished 5/9 of the 45 math problems he was assigned. If each math problem took him 3/5 of a minute to finish an

Oliga [24]

Step-by-step explanation:

plz refer the attachment

6 0

3 years ago

What is the solution of 3+(x-2/x-3)<=4

Minchanka [31]

Answer: x < 3

Step-by-step explanation:

Solve the rational inequality by combining expressions and isolating the variable x.

7 0

3 years ago

Which function has the greater maximum value: f(x) = -2x2 + 8x-1, or g(x),

aleksklad [387]

Answer:

It’s f(x)

Step-by-step explanation:

4 0

3 years ago

The steps to solving the expression the quantity of the sum of three squared plus four divided by the absolute value of negative

bearhunter [10]

Answer:

Step 3 should be ((9+4)/(0.25))-10.4 divided 4

Step 6 should be 52 - 2.6

I got it right on my exam :)

7 0

1 year ago

If <img src="https://tex.z-dn.net/?f=tan%20%28x%29%20%3D%20%5Cfrac%7B5%7D%7B12%7D" id="TexFormula1" title="tan (x) = \frac{5}{12

Alekssandra [29.7K]

Explanation:

First, we need to find the values of the sine and cosine of x knowing the value of tan x and x being in the 3rd quadrant. Since tan x = 5/12, using Pythagorean theorem, we know that

$\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}$

Note that both sine and cosine are negative because x is in the 3rd quadrant.

Recall the addition identities listed below:

$\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta$

$\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x$

$\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}$

$\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$

$\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x$

$\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}$

$\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$

$\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}$

$\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}$

Therefore, the expression reduces to

$\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}$

$\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}$

$\;\;\;\;\;= \dfrac{49}{26}$

5 0

2 years ago