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3 years ago

Evalute the expression. 2(2+4)^0+3(18+6)^0 A) 0 B) 1 C) 2 D) 5

Mathematics

2 answers:

EastWind [94]3 years ago

7 0

Answer:

$\boxed{\bold{5}}$

Step-by-step explanation:

Follow PEMDAS Order Of Operations

(2 + 4): 6

$\bold{2\cdot \:6^0+3\left(18+6\right)^0}$

(18 + 6): 24

$\bold{2\cdot \:6^0+3\cdot \:24^0}$

$\bold{6^0: \ 1}$

$\bold{2\cdot \:1+3\cdot \:24^0}$

$\bold{24^0: \ 1}$

$\bold{2\cdot \:1+3\cdot \:1}$

$\bold{2\cdot \:1: \ 2}$

$\bold{2+3\cdot \:1}$

$\bold{3\cdot \:1: \ 3}$

$\bold{2+3}$

$\bold{2+3: \ 5}$

Anastaziya [24]3 years ago

6 0

Answer:

D, 5

Step-by-step explanation:

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kati45 [8]

The answer is (2, 0).

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2 years ago

A baseball player comes up to bat 3 times during a league game. He either gets a hit or gets an out. How many different combinat

Stella [2.4K]

Answer:

8 different combinations are possible.

Step-by-step explanation:

Here, we have 2 different combinations for each time.

And the player comes out to bat 3 times.

So, total number of combinations are:

$2^{3}$ i.e. a total of 8 number of times.

Let a hit is termed as 'H' and an out is termed as 'O'.

Total combinations are:

{HHH, HHO, HOH, HOO, OHH, OHO, OOH, OOO}

Kindly have a look at the tree diagram attached in the answer area.

In starting, there are 2 combinations possible, i.e. 'O' and 'H'.

After 'O' , 2 possible i.e. 'O' and 'H'.

After 'H' , 2 possible i.e. 'O' and 'H'.

and so on....

8 0

3 years ago

(◞ ‸ ◟) PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP ≧﹏≦

Natasha_Volkova [10]

6. Classify the angle as acute, right, obtuse, or straight.

A. obtuse

B. acute

C. straight

D. right

(THE ANSWER IS OBTUSE ANGLE)

An Obtuse Angle Is what your looking at right now. Its angle is bigger than 90 degree.

A Acute angle is lower than 90 degree which tends to make something looking like the “<” “>” sign.

And straight angle is just a straight line (No kidding)

A right angle is justs a L shape sign. The corner of the L 90 degree

3 0

2 years ago

A discus is thrown from a height of 3 feet with an initial velocity of 55 ft/s at an angle of 44° with the horizontal. How long

nasty-shy [4]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Yes, the 3 feet are important.
You can consider the datum (y=0) at 3' from the ground, and the ground is therefore at -3 feet. 

The vertical initial velocity, vy, is u*sin(θ), so it gives 
vy=55sin(44°)=38.21 approx. 

The (vertical) distance travelled is given by: 
S=vy*t-(1/2)gt² 
where 
S=-3 (ground) 
vy=55sin(44°), and 
g=32.2 ft/s² 
Solve for t. 
I get -0.08 and 2.45 s.

6 0

3 years ago

Prove it please answer only if you know

deff fn [24]

Part (c)

We'll use this identity

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\$

to say

$\sin(A+45) = \sin(A)\cos(45) + \cos(A)\sin(45)\\\\\sin(A+45) = \sin(A)\frac{\sqrt{2}}{2} + \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\$

Similarly,

$\sin(A-45) = \sin(A + (-45))\\\\\sin(A-45) = \sin(A)\cos(-45) + \cos(A)\sin(-45)\\\\\sin(A-45) = \sin(A)\cos(45) - \cos(A)\sin(45)\\\\\sin(A-45) = \sin(A)\frac{\sqrt{2}}{2} - \cos(A)\frac{\sqrt{2}}{2}\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

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

The key takeaways here are that

$\sin(A+45) = \frac{\sqrt{2}}{2}(\sin(A)+\cos(A))\\\\\sin(A-45) = \frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\$

Therefore,

$2\sin(A+45)*\sin(A-45) = 2*\frac{\sqrt{2}}{2}(\sin(A)+\cos(A))*\frac{\sqrt{2}}{2}(\sin(A)-\cos(A))\\\\2\sin(A+45)*\sin(A-45) = 2*\left(\frac{\sqrt{2}}{2}\right)^2\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = 2*\frac{2}{4}\left(\sin^2(A)-\cos^2(A)\right)\\\\2\sin(A+45)*\sin(A-45) = \sin^2(A)-\cos^2(A)\\\\$

The identity is confirmed.

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

Part (d)

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\\\\\sin(45+A) = \sin(45)\cos(A) + \cos(45)\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}\cos(A) + \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\$

Similarly,

$\sin(45-A) = \sin(45 + (-A))\\\\\sin(45-A) = \sin(45)\cos(-A) + \cos(45)\sin(-A)\\\\\sin(45-A) = \sin(45)\cos(A) - \cos(45)\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}\cos(A) - \frac{\sqrt{2}}{2}\sin(A)\\\\\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\$

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

We'll square each equation

$\sin(45+A) = \frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\\\\\sin^2(45+A) = \left(\frac{\sqrt{2}}{2}(\cos(A)+\sin(A))\right)^2\\\\\sin^2(45+A) = \frac{1}{2}\left(\cos^2(A)+2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

and

$\sin(45-A) = \frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\\\\\sin^2(45-A) = \left(\frac{\sqrt{2}}{2}(\cos(A)-\sin(A))\right)^2\\\\\sin^2(45-A) = \frac{1}{2}\left(\cos^2(A)-2\sin(A)\cos(A)+\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\frac{1}{2}*2\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

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

Let's compare the results we got.

$\sin^2(45+A) = \frac{1}{2}\cos^2(A)+\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\\sin^2(45-A) = \frac{1}{2}\cos^2(A)-\sin(A)\cos(A)+\frac{1}{2}\sin^2(A)\right)\\\\$

Now if we add the terms straight down, we end up with $\sin^2(45+A)+\sin^2(45-A)$ on the left side

As for the right side, the sin(A)cos(A) terms cancel out since they add to 0.

Also note how $\frac{1}{2}\cos^2(A)+\frac{1}{2}\cos^2(A) = \cos^2(A)$ and similarly for the sin^2 terms as well.

The right hand side becomes $\cos^2(A)+\sin^2(A)$ but that's always equal to 1 (pythagorean trig identity)

This confirms that $\sin^2(45+A)+\sin^2(45-A) = 1$ is an identity

4 0

3 years ago