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4 years ago

Give an example of a quadratic with complex roots. Explain how you know it has complex roots.

1 answer:

5 0

2x^2 + x + 3=0 has only complex roots.

The determinant is 1-4*2*3 = -23

-23 (or any determinant) is the part under the square root sign, If that determinant is negative, knowing you cannot take the square root of a negative number, we know the answers must be complex.

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The formula of the area of the triangle is A=1/2bh solve for h

inn [45]

Answer:

Answer:

h = \frac{2A}{b}

Step-by-step explanation:

Given

A = \frac{1}{2} bh

Multiply both sides by 2 to clear the fraction

2A = bh ( isolate h by dividing both sides by b )

h = \frac{2A}{b}

4 0

3 years ago

Can you help please ASAP

V125BC [204]

Answer:

Step-by-step explanation:

External angle equals to the sum of opposite internal angles

z - 41° + z - 20° = z + 40°

2z - 61° = z + 40°

2z - z - 61° = 40°

z - 61° = 40

z = 40° + 61°

z = 101°

5 0

3 years ago

a collection of dimes and quarters is worth $19.85. There are 128 coins in all. How many of each type of coin are in the collect

PolarNik [594]

Number of dimes were 81 and number of quarters were 47

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that There are 128 coins in all

number of dimes + number of quarters = 128

d + q = 128 ------ eqn 1

Also given that collection of dimes and quarters is worth $19.85

number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85

$d \times 0.10 + q \times 0.25 = 19.85$

0.1d + 0.25q = 19.85 -------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 128 - q -------- eqn 3

Substitute eqn 3 in eqn 2

0.1(128 - q) + 0.25q = 19.85

12.8 - 0.1q + 0.25q = 19.85

12.8 + 0.15q = 19.85

0.15q = 7.05

Therefore from eqn 3,

d = 128 - q

d = 128 - 47

Thus number of dimes were 81 and number of quarters were 47

4 0

3 years ago

Identify the vertex for the graph of y = −3x^2 + 6x − 4

algol [13]

Answer:

the vertex is (1,-1)

Step-by-step explanation:

y = −3x^2 + 6x − 4

the vertex is at

h= -b/2a where ax^2 +bx+c

h= -6/2(-3)

h= -6/-6

h=1

substitute this in to find the y value

y = -3(1)^2 +6(1)-4

y = -3+6-4

y = -1

the vertex is (1,-1)

3 0

3 years ago

I need help please!!!

V125BC [204]

Answer:

all you gotta do is convert miles to yards using google. (or bing if you are a nerd)

Step-by-step explanation:

8 0

3 years ago

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