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3 years ago

What is the modulus of |9+40i|?

Mathematics

2 answers:

Gekata [30.6K]3 years ago

8 0

Answer: Modulus of |9+40i| =41

Step-by-step explanation:

Since we have given that

$z=\mid9+40i\mid$

Here,

$z=a+bi\\\\where,\\\\a=\text{real number}=9\\\\b=\text{imaginary number}=40$

We need to find the modulus of z:

$\mid z\mid=\sqrt{a^2+b^2}$

$\mid z\mid=\sqrt{9^2+40^2}\\\\\mid z\mid=\sqrt{81+1600}\\\\\mid z\mid=\sqrt{1681}\\\\\mid z\mid=41$

Hence, Modulus of |9+40i| =41

Talja [164]3 years ago

6 0

Answer:

Step-by-step explanation:

We know that complex numbers are a combination of real and imaginary numbers

Real part is x and imaginary part y is multiplied by i, square root of -1

Modulus of x+iy = $\sqrt{x^2+y^2}$

Here instead of x and y are given 9 and 40

i.e. 9+40i

Hence to find modulus we square the coefficients add them and then find square root

|9+49i| = $\sqrt{9^2+40^2} =\sqrt{1681}$

By long division method we find that

|9+40i| =41

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Read 2 more answers

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Lera25 [3.4K]

Answer:

$130.2845\leq h\leq 137.7245$

Step-by-step explanation:

Given an inequality that relates the height h, in centimeters, of an adult female and the length f, in centimeters, of her femur by the equation

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$h - 134.0045 \leq 3.72\\h \leq 3.71+134.0045\\h\leq 137.7245$

If the modulus function is negative then:

$-(h - 134.0045) \leq 3.72\\-h+134.0045 \leq 3.72\\-h\leq 3.72-134.0045\\-h\leq -130.2845\\$

multiply through by -1

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combining the resulting inequalities, the estimate of the possible range of heights will be $130.2845\leq h\leq 137.7245$

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