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3 years ago

P(8,-3) T(x,y) = (x-4, y+7) P'(?,?)

Mathematics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

P'(4,4)

Step-by-step explanation:

P(8,-3) T(x,y) = (x-4, y+7)

P'(8-4,-3+7)=P'(4,4)

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2 years ago

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Answer:

Option (b) is correct.

$(2^\frac{1}{4} )^4=2^\frac{1}{4} \times 2^\frac{1}{4}\times 2^\frac{1}{4}\times 2^\frac{1}{4}=2^{(\frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} )}=2^{1}=2$

Step-by-step explanation:

Given: $(2^\frac{1}{4} )^4$

We have too choose the correct simplification for the given statement.

Consider $(2^\frac{1}{4} )^4$

Using property of exponents, $(a^m)^n=a^m\times a^m\times a^m\times ....\times (n\ times)$

We have,

$(2^\frac{1}{4} )^4=2^\frac{1}{4} \times 2^\frac{1}{4}\times 2^\frac{1}{4}\times 2^\frac{1}{4}$

Again applying property of exponents $a^m\times a^m=a^{n+m}$

We have,

$(2^\frac{1}{4} )^4=2^{(\frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} )}$

Simplify, we have,

$(2^\frac{1}{4} )^4=2^{\frac{4}{4}}$

we get,

$(2^\frac{1}{4} )^4=2^{1}=2$

Thus, $(2^\frac{1}{4} )^4=2$

Option (b) is correct.

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