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1 year ago

The operator *= is called a(n) _____ operator.

Mathematics

1 answer:

timama [110]1 year ago

7 0

The operator *= is called multiple assignment operator.

<h3>How to determine the name of the operator?</h3>

The operator is given as

Take for instance, we have the following statement in Python

a *= b

The meaning of the above statement is

a = a * b

Assume the following values for the variables

a = 6 and b = 3

The value of the operation a *= b would be

a *= b --- given expression

a = a * b --- expanded expression

a = 6 * 3 ---- when the values are substituted

a = 18 --- the result

This implies that the operator represents an assignment and a multiply operator

Hence, the operator *= is called multiple assignment operator.

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18. You are standing beside the Colorado River, 400 feet

Veseljchak [2.6K]

<h3>Answer: B. 781.6 feet approximately</h3>

=======================================================

Work Shown:

The horizontal portion is 400+166 = 566 feet. Label this as 'a', so a = 566. The vertical side is unknown, so b = x. The hypotenuse is c = 965

Use the pythagorean theorem

a^2+b^2 = c^2

566^2+x^2 = 965^2

x^2 = 965^2 - 566^2

x = sqrt( 965^2 - 566^2 )

x = 781.58108984289 which is approximate

x = 781.6 feet when rounding to one decimal place

6 0

3 years ago

F(x) = 9-3x<br> g(x) = 5x-7<br> Find f(x)+g(x).

Bezzdna [24]

Answer:

In the problem, the sum of the two functions is 2x + 2

Step-by-step explanation:

For this problem, we have to add together f(x) and g(x).

(f + g)(x) = (9 - 3x) + (5x - 7)

Combine like terms.

(f + g)(x) = 2x + 2

So, when you combine the two functions together, you will get 2x + 2.

8 0

3 years ago

Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0