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3 years ago

How do I solve (7-6i)(-8+3i)?

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

6 0

I = {2.666666667, 1.166666667

konstantin123 [22]3 years ago

5 0

Solution step by step

(7-6i)(-8+3i)
=7×(-8)+7×3i-6i×(-8)-6i×3i
=-56+21i+48i+18i^2
=-56+69i-18×(-1)
=-56+69i+18
=-38+39i

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When Lin drives a golf ball, it travels a mean distance of 200 yards with a standard deviation of 20 yards, and the distribution

Anettt [7]

Answer:

The percentage of her shots that will travel less than 180 yards is approximately 15.9%

Step-by-step explanation:

First it will be required to find the standard or z-score from the following formula

$z = \frac{x- \mu}{\sigma}$

Where:

z = z or standard score

x = value that is observed = 180 yards

μ = Sample mean = 200 yards

σ = Sample standard deviation yards

Plugging the values, we have;

$z = \frac{180- 200}{20} = -1$

Therefore, the probability that her shots will travel less than 180 is given by the relation;

p( x < 180 yards) = p(z < -1) = 0.15866

Therefore, the percentage of her shots that will travel less than 180 = 0.15866 × 100

= 15.866% ≈ 15.9%.

3 0

3 years ago

PLEASE HELP ASAP!!! i dont understand :(

aivan3 [116]

Answer:

hello,

Step-by-step explanation:

y=k(x-a')²+b'

focus is (a',1/(4k)+b')

2x²=-120y

==> y=-2/120x²

==> y=-1/60(x-0)²+0

focus= (0,-15)

7 0

3 years ago

Please please please need help on 3 and 4 I’m really lost on them

Alex73 [517]

Answer:

1) Option c) is correct ie., 5 real and o non-real

2) Option b) is correct ie., (4, $\frac{1}{2}$ , $\frac{1}{2}$ , 2,2)

Step-by-step explanation:

Given polynomial function is $f(x)=x^5-3x^3-2$

To find zeros equate f(x) to zero ie., $f(x)=0$

$x^5-3x^3-2=0$

By synthetic division

| 1 0 -3 0 -2

-1 | 0 -1 1 2 2

|_________________

1 -1 -2 2 0

Therefore x=-1 is a zero

$x^3-x^2-2x+2=0$

| 1 -1 -2 2

1 | 0 1 0 2

|___________________

1 0 -2 0

x=1 is the zero

$x^2 -2 =0$

$x=\pm\sqrt{2}$

$x=\sqrt{2}$ and

$x=\sqrt{-2}$

Option c) is correct ie., 5 real and o non-real

2) Given polynomial function is $f(x)=(x-4)(2x-1)^2(x-2)^2$

To find zeros equate f(x) to zero ie., $f(x)=0$

$(x-4)(2x-1)^2(x-2)^2=0$

$(x-4)=0$ (or) $(2x-1)^2=0$ or $(x-2)^2=0$

Therefore x=4, $x=\frac{1}{2}$ of multiplicity of 2 and x=2 multiplicity of 2

Option b) is correct ie., (4, $\frac{1}{2}$ , $\frac{1}{2}$ , 2,2)

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4 years ago

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WITCHER [35]

Answer:

q = -5 $\sqrt{5}$

q = 5 $\sqrt{5}$

Step-by-step explanation:

Since there is no constant value. This must be a special case.

So,

(q+5 $\sqrt{5}$ )(q-5 $\sqrt{5}$ )

q = -5 $\sqrt{5}$

q = 5 $\sqrt{5}$

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