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Strike441 [17]
3 years ago
11

I’ve been stuck on this for a min now

Mathematics
1 answer:
dedylja [7]3 years ago
4 0

Answer:

3/14

Step-by-step explanation:

1st subtract the y's

the subtract the x's in the same order

-9 - 5 = -14

put the first result over the second

-3/-14 = 3/14

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Find the area of the triangles round to the tenth. Show steps please!
amm1812

Answer:

\large\boxed{A=89.6}

Step-by-step explanation:

It's the right triangle. The formula of an area of a right triangle:

A=\dfrac{leg\cdit leg}{2}

We have:

leg=x\\leg=16\\\text{and the angle}\ \alpha=55^o

Use tangent:

tangent=\dfrac{opposite}{adjacent}

opposite=16,\ adjacent=x

\tan55^o\approx1.4281

Substitute:

\dfrac{16}{x}=1.4281        <em>convert the decimal to the fraction</em>

\dfrac{16}{x}=\dfrac{14281}{10000}          <em>cross multiply</em>

14281x=160000            <em>divide both sides by 14281</em>

x=\dfrac{160000}{14281}

x\approx11.2

Calculate the area:

A=\dfrac{(16)(11.2)}{2}=89.6

7 0
3 years ago
Factor this expression as a difference of squares: 4-(2a+3b)^2
Nadya [2.5K]
The complete factorized expression is (2 + 2a + 3b) (2 - 2a + 3b).
Take a look at the attachment to see how I got this solution. 

7 0
3 years ago
Find an identity for cos(4t) in terms of cos(t)
mestny [16]
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.

Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1

Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.

Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.

Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1

2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute

8Cos^4 (t) -8Cos^2 (t) +1

Cos (4t) =8Cos^4-8Cos^2 (t)+-1






5 0
3 years ago
How many different arrangements are there of the letters a, b, c, d, e, f for which e is not last in line?
Olegator [25]
Hmm, there are 5 letters


if you cannot repeat the letters
hmm, let's find how many for 4 letters
there are 4! or 4*3*2*1=24
that's how many ways there are to arrange a,b,c, d
now we can multiply it by 4 because there are 4 choices for the last place (e is not included0
24*4=56
answer is 56 different arrangements
6 0
3 years ago
Lamar bought a table on sale for $399. This price was 24% less than the original price. What was the original price?
xxMikexx [17]

$525

Hope this helps ya out

4 0
3 years ago
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