4 years ago

How can you use 4x7 to find 8x7

Mathematics

1 answer:

CaHeK987 [17]4 years ago

5 0

Hello!
4x7=28
and
8x7=56
28+28=56
Hope this helped!

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Answer:

1/2, 1/4 & 3/4

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mean of 0 &1 is $\frac{0+1}{2} =\frac{1}{2} \\$

mean of 0 and 1/2 is $(0+\frac{1}{2} )/2=\frac{1}{2} /2=\frac{1}{2} *\frac{1}{2} =\frac{1}{4}$

mean of 1/2 and 1 is $(\frac{1}{2} +1)/2=\frac{3}{2} /2=\frac{3}{2} *\frac{1}{2} =\frac{3}{4}$

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galina1969 [7]

How to get answer for number 1: | 4+2i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=4,\:b=2\\=\sqrt{4^2+2^2}\\Refine\\=\sqrt{20}\\\sqrt{20}=2\sqrt{5}\\=2\sqrt{5}$

How to get answer for number 2: | 5-i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=5,\:b=-1\\=\sqrt{5^2+\left(-1\right)^2}\\=\sqrt{26}$

Number 3 how to get answer: | -3i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=0,\:b=-3\\=\sqrt{0^2+\left(-3\right)^2}\\Refine\\=\sqrt{9}\\\sqrt{9}\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=\sqrt{3^2}\\\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a\\\sqrt{3^2}=3\\= 3$