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jeyben [28]
4 years ago
6

Find all complex solutions of 3x^2+2x+5=0 . (If there is more than one solution, separate them with commas.)

Mathematics
1 answer:
barxatty [35]4 years ago
8 0

Answer:

\large\boxed{-\dfrac{1}{3}-\dfrac{\sqrt{14}}{3}i,\ -\dfrac{1}{3}+\dfrac{\sqrt{14}}{3}i}

Step-by-step explanation:

Use

ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

3x^2+2x+5=0\\\\a=3,\ b=2,\ c=5\\\\b^2-4ac=2^2-4(3)(5)=4-60=-56\\\\\sqrt{b^2-4ac}=\sqrt{-56}=\sqrt{(4)(-14)}=\sqrt4\cdot\sqrt{-14}=2\sqrt{-14}

Use

i=\sqrt{-1}

\sqrt{-14}=\sqrt{(-1)(14)}=\sqrt{-1}\cdot\sqrt{14}=i\sqrt{14}

Therefore:

x=\dfrac{-2\pm 2i\sqrt{14}}{2(3)}=-\dfrac{2}{6}\pm\dfrac{2i\sqrt{14}}{6}=-\dfrac{1}{3}\pm \dfrac{\sqrt{14}}{3}i

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What are the zeros for the function
Greeley [361]

The zeroes of the function are x = -3, \ \ x =2\sqrt{ 3}i,  \ \ x =-2\sqrt{ 3}i.

Solution:

Given function:

f(x)=x^{3}+3 x^{2}+12 x+36

<u>To find the zeros of the function:</u>

⇒ f(x) = 0

\Rightarrow x^{3}+3 x^{2}+12 x+36=0

\Rightarrow (x^{3}+3 x^{2})+(12 x+36)=0

Take x² as common in first bracket and take 12 as common in 2nd bracket.

\Rightarrow x^2(x+3 )+12( x+3)=0

Now, take common term (x + 3) outside.

\Rightarrow (x+3 ) (x^2+12)=0

<em>Using zero factor principle, If ab = 0 then a = 0 or b = 0.</em>

x+3=0,  \ \ x^2+12=0

x = –3

x^2+12=0

x² = –12

x² = 2² × 3 × –1

Taking square root on both sides, we get

\sqrt{x^2} =\pm\sqrt{2^2\times 3\times -1}

x =\pm2\sqrt{ 3\times -1}

we know that \sqrt{-1} =i.

x =\pm2\sqrt{ 3}i

x =2\sqrt{ 3}i,  \ \ x =-2\sqrt{ 3}i

Hence the zeroes of the function are x = -3, \ \ x =2\sqrt{ 3}i,  \ \ x =-2\sqrt{ 3}i.

7 0
3 years ago
The midpoints of the hexagon are connected to form another hexagon, this pattern continues indefinitely. If the area of the 99th
jarptica [38.1K]

Answer:

  • (4/3)⁹⁸ units²

Step-by-step explanation:

  • <em>Regular hexagons have a property that cutting off the triangles obtained by joining the midpoints of consecutive sides leaves a hexagon of 3/4 of the area.</em>

This means the area of each hexagon inside out starting from the one with unit area is 4/3 of the previous one.

  • 99 ⇒ 1
  • 98 ⇒ 1*4/3
  • 97 ⇒ 1*(4/3)²
  • ...
  • 1 ⇒ (4/3)⁹⁹⁻¹

<u>The original hexagon has the area of:</u>

  • (4/3)⁹⁸ units²
7 0
3 years ago
Read 2 more answers
The average age of Iowa residents is 37 years. Amy believes that the average age in her hometown in Iowa is not equal to this av
alekssr [168]

Answer:

The p-value of the test is 0.2.

Step-by-step explanation:

Using the alternative hypothesis that µ ≠ 531, Amy found a t-test statistic of 1.311.

Alternative hypothesis means that e have a two-tailed hypothesis test.

Sample 30 citizens. Amy found a t-test statistic of 1.311.What is the p-value of the test statistic?

Thus, we have a two-tailed test with 30 - 1 = 29 degrees of freedom and t = 1.311. So, using a t-distribution calculator, the p-value of the test is 0.2.

6 0
3 years ago
I need help on this one question please someone help me pleases
Masteriza [31]
So we simplify this and do
5(3a-1)-2(3a+2)=3(a+2)+v
15a-5-6a-4=3a+6+v
add like terms
9a-9=3a+6+v
subtract 6 from both sides and get
9a-15=3a+v
subtract 3a from both sides and get
6a-15=v or 3(2a-5)
so the answers are D,E

or if you want to avoid doing allot of math, you could just figure out which two answer choices are the same (D,E)
4 0
3 years ago
Read 2 more answers
Pls show work! questions on pic
Lena [83]

Answer:

a.(10,23)

b.(7,30)

e.(20,20)

Step-by-step explanation:

We are given that

x=Width of garden

y=Length of garden

According to question

x\geq 5

y\geq 20

Perimeter of rectangle=2(x+y)

Where x=Width  of rectangle

y=Length of rectangle

Fencing used =2(x+y)

2(x+y)\leq 80

First we change inequality  into equality

x=5...(1)

y=20...(2)

2(x+y)=80

x+y=40...(3)

Substitute x=0 in equation (3)

y=40

Substitute y=0  in equation (3)

x=40

Substitute x=0 and y=0 in equation (3)

2(x+y)\leq 80

0\leq 80

It is true. Therefore, the  shaded region below the line.

0\geq 5

0\geq 20

These are false equation .Therefore, the shaded region above the line.

Substitute x=5 in equation (3)

5+y=40

y=40-5=35

Substitute y=20 in equation (3)

x+20=40

x=40-20=20

Intersection point of equation (1) and (3) is (5,35) and intersect point of equation (2) and (3) is (20,20).

Now,

a.(10,23)

Substitute in

2(x+y)\leq 80

2(10+23)\leq 80

66\geq 80

10>5 and 23>20

it satisfied all inequality.

Hence, it is a solution.

b.(7,30)

7>5

30>20

2(7+30)=74

it satisfied all inequality.

Hence, it is a solution.

c.(18,25)

18>5

25>20

2(18+25)=86>80

It does not satisfied third inequality

Hence, it is not a solution.

d.(8,35)

8>5

35>20

2(8+35)=86>80

It does not satisfied third inequality

Hence, it is not a solution.

e.(20,20)

2(20+20)=80

20>5

20=20

it satisfied all inequality.

Hence, it is a solution.

4 0
3 years ago
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