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storchak [24]
3 years ago
13

5x^2 + 19x + 12=0 how do i figure this out

Mathematics
1 answer:
zlopas [31]3 years ago
3 0

Answer:

x=−4/5 or x=−3

Step-by-step explanation:

Step 1: Factor left side of equation.

(5x+4)(x+3)=0

Step 2: Set factors equal to 0.

5x+4=0 or x+3=0

x=−4/5 or x=−3

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A magazine can layout 1/16 of an issue in 3 days. How many days does it take to layout one issue?
Zarrin [17]
Answer

48 Days


Explanation

3(days) times 16= 48 days
4 0
3 years ago
Pythagoras was born about 582 bc. Isaac Newton was born in 1643 ad. How many years apart were they born?
malfutka [58]
B.C. counts down to 0, and that's when A.D. starts counting up. That means you can add 582 and 1643, to get their difference. Pythagoras and Newton were born 2,225 years apart.
8 0
4 years ago
30% × 750 =<br> (30 ÷ 100) × 750 =<br> (30 × 750) ÷ 100 =<br> 22,500 ÷ 100 =<br> 225;
Deffense [45]

Answer:

SOLVE IT UR SELF OR GET A CALCULATOR

Step-by-step explanation:

*urinates on ur property*

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7 0
3 years ago
Read 2 more answers
Help please thank you
jek_recluse [69]

Answer:

the answer 1.5 if you want to know the answer see what the pattern is

4.5 - 3 = 1.5

so its going up by 1.5  :D

7 0
3 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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