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irakobra [83]
3 years ago
14

Can someone solve this?

Mathematics
1 answer:
Lelu [443]3 years ago
7 0
4) 18*24+ 7^2*(22/7)/2 = 432*77=509cm^2
5) 5*2*15 + 5^2*3.14=228.5yd^2
6)31*24/2+16^2 =372+ 256 =628m^2
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50 Pts!! Brainliest!! Answer ASAP, thx.
Y_Kistochka [10]

Answer:

1/625

Step-by-step explanation:

When you multiply, you add the exponents. You now have 6^{-4}. This is equivalent to \frac{1}{5^{4} } which equals 1/625.

6 0
3 years ago
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i have been trying to solve this compound and double angle question please help me find the answer to these question guys​
natali 33 [55]

Answer:

These type of questions are super tricky b/c you have to remember all the different versions of the identities, and then they put the question in some odd form,  I feel like this should land math professors in jail , for dishonesty , b/c it's really a form of "how tricky can I make a question and still have a way to solve it"   anyway,

Step-by-step explanation:

a)

next the question asks   1-cos 2A   and this is total abuse of notation.   the way this should be written is   1- cos( 2A)  so we know that the A is part of the cosine functions input... btw.. in any computer program,  it would never ever let you get away with that top form of the expression.  :/   anyway... I keep ranting.. huh... sorry  :P

1-cos(2A) is an odd form of the identity  1/2(1-cos(2A) = sin^{2}(A)  the 1/2 is missing but we can add that pretty easy, we just have to remember to take it out too. I usually forget to do that. and my professor marks me off completely,  totally wrong, but I just miss one small thing  :/  anyway....

our 1-cos(2A) needs the 1/2 added to it.  or if we move that 1/2 to the other side it looks like  2*sin^{2}(A)  = 1-cos(2A)  and this is that "odd" from of the identity that I was talking about.  

next let's deal with sin(2A)  it has an identity of  2 sin(A)cos(A) which is really nice for us b/c it will cancel out the 2 in then numerator for us, nice !

now our fraction looks like  [2* sin^{2}(A)] / 2 sin(A)cos(A)

so cancel out one of the sines

2*sin(A) / 2 cos(A)

cancel the 2s

Sin(A) / Cos(A) = Tan(A)

nice  it worked out  :P

b)

by the above that we just worked out, then

Tan(15) = Sin(15) / Cos(15)

I had to look up what sin of 15 is b/c it's not one of those special angles but it does have an exact form of

Sin(15) = (√3 - 1) / 2√2

Cos(A) = (√3 + 1) / 2√2

you can use rule of Cos(A-B) = Cos(A)Cos(B)+Sin(A)Sin(B) to get the above and a similar rule for Sin(A-B)

back to our problem,  the 2√2 will cancel out

then we have

Tan(15) =  (√3 - 1) /(√3 + 1)

in the form that is above that's exact, the roots could be approximated but i'll just leave that in the form that is exact.  Most math professors like that form.  

 

4 0
2 years ago
Suppose you had d dollars in your bank account. You spent $21 but have at least $53 left. How much money did you have initially?
likoan [24]
D-21≥ 53 hope this helps
6 0
3 years ago
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Jasmine sold 90 rolls of wrapping paper.Jasmine sold 2times as many wrapping paper as Carly.How many rolls of wrapping paper did
hodyreva [135]
135 because if you divide 90 by 2 you get 45 then add 45 and 90 and you get 135
5 0
3 years ago
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Screenshot of the question attatched. please help me
Law Incorporation [45]
<h2>Rules of Exponents</h2>

There are many rules that apply to operations involving exponents. In this question, we'll need the following:

<h3>The division rule</h3>
  • ⇒ \dfrac{a^n}{a^m}=a^{n-m}

<h2>Solving the Question</h2>

We're given:

z^{21}\div z^7=z^k (solve for k.)

⇒ Rewrite as a fraction:

\dfrac{z^{21}}{z^7}=z^k

⇒ Use the division rule:

z^{21-7}=z^k\\z^{14}=z^k\\14=k

Therefore, the value of <em>k</em> is 14.

<h2>Answer</h2>

<em>k</em> = 14

5 0
2 years ago
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