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Nataly [62]
3 years ago
5

Solve This Using The Quadratic Formula: 3x^2+9x-6=0

Mathematics
1 answer:
aivan3 [116]3 years ago
4 0
<h3>Therefore either  x = \frac{-3+\sqrt{17}}{2}    or,  x = \frac{-3-\sqrt{17}}{2}</h3>

Step-by-step explanation:

3x^2 +9x -6=0                                x = \frac{-b\pm\sqrt{b^2- 4ac} }{2a}

\Leftrightarrow x = \frac{-9\pm\sqrt{9^2- 4.3(-6)} }{2.3}                     here a = 3 ,b = 9 and c= -6

\Leftrightarrow x = \frac{-9\pm\sqrt{153} }{6}

\Leftrightarrow x = \frac{3(-3\pm\sqrt{17}) }{6}

\Leftrightarrow x = \frac{-3\pm\sqrt{17}}{2}

Therefore either  x = \frac{-3+\sqrt{17}}{2}    or,  x = \frac{-3-\sqrt{17}}{2}

You might be interested in
HELP ASAP (Geometry)
Andrei [34K]

1) Parallel line: y=-2x-3

2) Rectangle

3) Perpendicular line: y = 0.5x + 2.5

4) x-coordinate: 2.7

5) Distance: d=\sqrt{(4-3)^2+(7-1)^2}

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: y-5=-4(x-(-1))

Step-by-step explanation:

1)

The equation of a line is in the form

y=mx+q

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

y=-2x+7

So its slope is m=-2. Therefore, the only line parallel to this one is the line which have the same slope, which is:

y=-2x-3

Since it also has m=-2

2)

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32

- Second diagonal: d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32

The diagonals are congruent, so this is a rectangle.

3)

Given points A (0,1) and B (-2,5), the slope of the line is:

m=\frac{5-1}{-2-0}=-2

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

m'=\frac{1}{2}

And using the slope-intercept for,

y-y_0 = m(x-x_0)

Using the point (x_0,y_0)=(7,1) we find:

y-1=\frac{1}{2}(x-7)

And re-arranging,

y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5

4)

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7

5)

The distance between two points A(x_A,y_A) and B(x_B,y_B) is given by

d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

d=\sqrt{(4-3)^2+(7-1)^2}

6)

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

\frac{3}{3+5}=\frac{3}{8}

of the  distance between A and B.

7)

To find the perimeter, we have to calculate the length of each side:

d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6

d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2

d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2

d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

8)

The area of a triangle is

A=\frac{1}{2}(base)(height)

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7

Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8

So the area is

A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8

9)

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

y=\frac{1}{4}x-6

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

m'=-4

Using the slope-intercept form,

y-y_0 = m'(x-x_0)

And using the point

(x_0,y_0)=(-1,5)

we find:

y-5=-4(x-(-1))

Learn more about parallel and perpendicular lines:

brainly.com/question/3414323

brainly.com/question/3569195

#LearnwithBrainly

8 0
3 years ago
What best describes how to evaluate a variable expression?
Kaylis [27]
<span> To replace each letter with its value, and then finish it by doing the order of the operation/ solve it.</span>
7 0
3 years ago
The question is in the image below.
sineoko [7]

Answer:

The Recursive Formula for the sequence is:

\:a_n=\frac{1}{5}\left(a_{n-1}\right)     ; a₁ = 125

Hence, option D is correct.

Step-by-step explanation:

We know that a geometric sequence has a constant ratio 'r'.

The formula for the nth term of the geometric sequence is

a_n=a_1\cdot r^{n-1}

where

aₙ is the nth term of the sequence

a₁ is the first term of the sequence

r is the common ratio

We are given the explicit formula for the geometric sequence such as:

a_n=125\left(\frac{1}{5}\right)^{n-1}

comparing with the nth term of the sequence, we get

a₁ = 125

r = 1/5

Recursive Formula:

We already know that

  • a₁ = 125
  • r = 1/5

We know that each successive term in the geometric sequence is 'r' times the previous term where 'r' is the common ratio.

i.e.

a_n=ra_{n-1}

Thus, substituting r = 1/5

\:a_n=\frac{1}{5}\left(a_{n-1}\right)  

and a₁ = 125.

Therefore, the Recursive Formula for the sequence is:

\:a_n=\frac{1}{5}\left(a_{n-1}\right)     ; a₁ = 125

Hence, option D is correct.

3 0
3 years ago
Jim is experimenting with a new drawing program on his computer. He created quadrilateral TEAM with coordinates T(- 2, 3), E(- 5
Crazy boy [7]

Answer: yes Jim is correct

5 0
2 years ago
Please help me it would help a lot
Juliette [100K]

Answer:

7/80 is the answer because there are 7 possible chances she can get a red or blue gumball.

Step-by-step explanation:

5 0
3 years ago
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