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2 years ago

Describe the nature of the roots for this equation. 3x^2+x-5= 0

Mathematics

2 answers:

Colt1911 [192]2 years ago

7 0

Answer:

two real,irrational roots.

GOOD LUCK!

Step-by-step explanation:

m_a_m_a [10]2 years ago

4 0

Answer:

<h2>This equation has no natural roots.</h2>

Step-by-step explanation:

$3x^2+x-5=0\\\\\text{Use the quadratic formula for}\ ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{For the given equation:}\\\\a=3,\ b=1,\ c=-5\\\\b^2-4ac=1^2-4(3)(-5)=1+60=61\\\\x=\dfrac{-1\pm\sqrt{61}}{(2)(3)}=\dfrac{-1\pm\sqrt{61}}{6}\notin\mathbb{N}$

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Phoenix [80]

Since the length is 15, and the width would be 15, you need to first do 15 x 15.

15 x 15 = 225

Multiply the 6 faces.

225x6

Your final answer would be 1,350.

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3 years ago

In an experiment, there are n independent trials. For each trial, there are three outcomes, A, B, and C. For each trial, the pro

Mekhanik [1.2K]

Answer:

(B) No. A binomial probability model applies to only two outcomes per trial.

Step-by-step explanation:

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3 years ago

Please help 70 Brainly points thanks

Cloud [144]

Answer:

9) x=58

10) r=4

11) m=4

12) p=3

13) x=6

14) x=-3

15) s=400

Step-by-step explanation:

9) x+2/5=12

x5 x5

x+2=60

-2 -2

x=58

10) 7r + 14 - 3r =30

-14 -14

<u>7r-3r</u>=16

4r=16

÷4 ÷4

r=4

11) $\frac{4}{5}$ m+2=6

-2 -2

$\frac{4}{5}$ m=4

÷ $\frac{4}{5}$ ÷ $\frac{4}{5}$

m=4

12) <u>2</u>(5p+9)=48

10p+18=48

-18 -18

10p=30

÷10 ÷10

p=3

13) <u>5</u>(2x-8)=20

10x-40=20

+40 +40

10x=60

÷10 ÷10

x=6

14)<u>6</u>(3-2x)=54

18-12x=54

-18 -18

-12x=36

÷-12 ÷-12

x=-3

15) $\frac{2s}{5}$ - $\frac{s}{2}$ =40

x5 x5

2s- $\frac{s}{2}$ =200

x2 x2

<u>2s-1s</u>=400

1s=400

÷1 ÷1

s=400

6 0

3 years ago

Read 2 more answers

How do I solve question 6 through 8?<br> Solve for me

rewona [7]

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

<h3>How to determine the functions?</h3>

A quadratic function is represented as:

y = a(x - h)^2 + k

<u>Question #6</u>

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

<u>Question #7</u>

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

3 = a(1 - 2)^2 + 1

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

<u>Question #8</u>

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

The graph pass through (0, -3)

So, we have:

-3 = a(0 - 1)^2 - 2

Evaluate the like terms

a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

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

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2