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Paladinen [302]
3 years ago
11

How to solve x^2-4x-5=0 by using the quadratic formula?

Mathematics
1 answer:
padilas [110]3 years ago
3 0

Answer:

x = 5 or -1, put numbers into the formula and calculate

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⚠WILL GET BRAINLEST⚠Select the table that represents a linear function. (Graph them if necessary.) ​
lara31 [8.8K]
A linear function is a trend that is equal to each other or corresponds to another set of numbers. The answer here would be C. because the multiples properly align in the table.
6 0
3 years ago
Write -3i + (3/4+2i) - (9/3+3i) as a complex number in standard form
bogdanovich [222]

Answer:

=  -3i + (3/4+2i) - (9/3+3i)

= (3/4 - 9/3) + (-3 + 2 -3)i

= -9/4 -4i

4 0
4 years ago
Find an equation of the line through the given point and perpendicular to the given line
s344n2d4d5 [400]

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

For two lines to be perpendicular, the slope of one line is the negative reciprocal of the other line. The equation of the given line is

y = 2x - 2

Comparing with the slope intercept form,

Slope, m = 2

This means that the slope of the line that is perpendicular to it is -1/2

The given points are (-3, 5)

To determine c,

We will substitute m = -1/2, y = 5 and x = - 3 into the equation, y = mx + c

It becomes

5 = -1/2 × - 3 + c

5 = - 3/2 + c

c = 5 + 3/2

c = 13/2

The equation becomes

y = -x/2 + 13/2

4 0
3 years ago
If 6xa=54 and b-a=14 what is axb
bagirrra123 [75]
6a=54 so a=54/6, a=9

b-a=14, and using a found above...

b-9=14

b=23,  so a=9 and b=23 so

ab=9(23)=207
6 0
3 years ago
Read 2 more answers
Evaluate the expression and write your answer in the form a + bi. 1) (2-6i)+(4+2i)2) (6+5i)(9-2i)3) 2/(3-9i)4) (3 − 5i)(7 − 2i)
laiz [17]

Answer:

Step-by-step explanation:

Given the following complex numbers, we are to expressed them in the form of a+bi where a is the real part and b is the imaginary part of the complex number.

1) (2-6i)+(4+2i)

open the parenthesis

= 2-6i+4+2i

collect like terms

= 2+4-6i+2i

= 6-4i

2)  (6+5i)(9-2i)

= 6(9)-6(2i)+9(5i)-5i(2i)

= 54-12i+45i-10i²

= 54+33i-10i²

In complex number i² = -1

= 54+33i-10(-1)

= 54+33i+10

= 54+10+33i

= 64+33i

3) For the complex number 2/(3-9i), we will rationalize by multiplying by the conjugate of the denominator i.e 3+9i

= 2/3-9i*3+9i/3+9i

=2(3+9i)/(3-9i)(3+9i)

= 6+18i/9-27i+27i-81i²

= 6+18i/9-81(-1)

= 6+18i/9+81

= 6+18i/90

= 6/90 + 18i/90

= 1/15+1/5 i

4) For (3 − 5i)(7 − 2i)

open the parenthesis

= 3(7)-3(2i)-7(5i)-5i(-2i)

= 21-6i-35i+10i²

= 21-6i-35i+10(-1)

= 21-41i-10

= 11-41i

4 0
3 years ago
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