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rewona [7]
3 years ago
11

How do I solve a 90° rotation about the origin

Mathematics
1 answer:
Dima020 [189]3 years ago
4 0
Here is a way my class does this. We basically copy the graph and points on some tracing paper (you can find this in the art shop if you have time) and rotate the tracing paper 90* counter or 90* clockwise and there is your answer :)
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Leno4ka [110]

Answer: x=\frac{37}{9}

Step-by-step explanation:

By the negative exponent rule, you have that:

(\frac{1}{a})^n=a^{-n}

By the exponents properties, you know that:

(m^n)^l=m^{(nl)}

(m^n)(m^l)=m^{(n+l)}

Rewrite 4, 8 and 32 as following:

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8=2³

32=2⁵

Rewrite the expression:

(2^2)^{(x-7)}*(2^3)^{(2x-3)}=\frac{32}{2^{(x-9)}}

Keeping on mind the exponents properties, you have:

(2)^{2(x-7)}*(2)^{3(2x-3)}=32(2^{-(x-9)}

(2)^{2(x-7)}*(2)^{3(2x-3)}=(2^5)(2^{-(x-9)})\\\\(2)^{(2x-14)}*(2)^{(6x-9)}=(2^5)(2^{(-x+9)})\\\\2^{((2x-14)+(6x-9))}=2^{(5+(-x+9))}

As the bases are equal, then:

(2x-14)+(6x-9)=5+(-x+9)\\\\2x-14+6x-9=5-x+9\\\\8x-23=14-x\\9x=37

x=\frac{37}{9}

3 0
4 years ago
Read 2 more answers
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