All categories

3 years ago

........ help ASAP please....

Mathematics

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

See below

Step-by-step explanation:

In the second step there should be (1 - cot x) instead of (1 + cot x)

(1 + tan x) [1 + cot(-x)]

= (1 + tan x) (1 - cot x)

= 1 - cot x + tan x - tan x cot x

= 1 - cot x + tan x - 1

= tan x - cot x

You might be interested in

Find the hypotenuse of a triangle with a base of 11cm and height of 9cm

netineya [11]

<span>Use Pythagorean Theorem :
</span><span>Hypotenuse
</span>c²=a²+b²
c²= √9²+11²
c=√202
c=14.21

7 0

3 years ago

For the last 4.5 hours, the temperature has decreased at a rate of 2.8 Degrees Fahrenheit per hour. Which best represents the ov

faust18 [17]

Answer:

The overall change in temperature over the time period ∆T = -12.6°F

Step-by-step explanation:

Rate of change of temperature r = -2.8°F per hour

Time t = 4.5 hours

The overall change in temperature ∆T = Rate of change of temperature × time period

∆T = r × t

∆T = -2.8°F per hour × 4.5 hours

∆T = -12.6°F

The overall change in temperature over the time period is -12.6°F

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

4 0

3 years ago

Identify the slope y-intercept of the function y= -2x + 3.

Anna11 [10]

Answer:

First option.

Step-by-step explanation:

Use the formula y = mx + b.

Where m is the slope, and b is the y-intercept.

y = -2x + 3

The slope is -2. The y-intercept is (0, 3).

5 0

3 years ago

Find angle D if angle B = 50

Mariana [72]

<h3>Answer: 80 degrees</h3>

============================================================

Explanation:

I'm assuming that segments AD and CD are tangents to the circle.

We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.

By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.

----------------------------

Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.

Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.

We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.

---------------------------

Here's what we have so far for quadrilateral DAEC

angle A = 90
angle E = 100
angle C = 90
angle D = unknown

Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees

A+E+C+D = 360

90+100+90+D = 360

D+280 = 360

D = 360-280

D = 80

Or a shortcut you can take is to realize that angles E and D are supplementary

E+D = 180

100+D = 180

D = 180-100

D = 80

This only works if AD and CD are tangents.

Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.

5 0

3 years ago