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musickatia [10]
3 years ago
7

Solve the quadratic equation 3x^2 – 5x – 7 = 0 Give your answers to 3 significant figures.

Mathematics
2 answers:
Oksana_A [137]3 years ago
6 0

Answer:

x≈2.57 or x≈-0.907

Step-by-step explanation:

plug the values a=3, b=-5 and c=-7 into

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

to find your answer

tankabanditka [31]3 years ago
5 0

Answer:

2.57 or -0.907

Step-by-step explanation:

→ First state the quadratic formula

\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

→ Identify the 'a', 'b' and 'c' variables

ax^{2} +bx+c

3x^{2} -5x-7

a = 3, b = -5 and c = -7

→ Substitute in the values into the quadratic formula

\frac{5+\sqrt{5^{2}-4*3*-7 } }{2*3}=\frac{5+\sqrt{25+84 } }{6}=\frac{5+\sqrt{109 } }{6}=2.573384418

→ Do the same for negative

\frac{5-\sqrt{5^{2}-4*3*-7 } }{2*3}=\frac{5-\sqrt{25+84 } }{6}=\frac{5-\sqrt{109 } }{6}=-0.9067177515

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Learning Task 3. Find the equation of the line. Do it in your notebook.
Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is y = 5\cdot x +9. The equation of the line in standard form is -5\cdot x + y = 9.

2) The equation of the line in slope-intercept form is y = \frac{2}{5}\cdot x +\frac{14}{5}. The equation of the line in standard form is -2\cdot x +5\cdot y = 14.

3) The equation of the line in slope-intercept form is y = 3\cdot x +4. The equation of the line in standard form is -3\cdot x +y = 4.

4) The equation of the line in slope-intercept form is y = 2\cdot x + 6. The equation of the line in standard form is -2\cdot x +y = 6.

5) The equation of the line in slope-intercept form is y = \frac{5}{6}\cdot x -\frac{7}{6}. The equation of the line in standard from is -5\cdot x + 6\cdot y = -7.

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: (m = 5, x = -1, y = 4)

4 = (5)\cdot (-1)+b

4 = -5 +b

b = 9

The equation of the line in slope-intercept form is y = 5\cdot x +9.

Then, we obtain the standard form by algebraic handling:

-5\cdot x + y = 9

The equation of the line in standard form is -5\cdot x + y = 9.

2) We begin with a system of linear equations based on the slope-intercept form: (x_{1} = 3, y_{1} = 4, x_{2} = -2, y_{2} = 2)

3\cdot m + b = 4 (Eq. 1)

-2\cdot m + b = 2 (Eq. 2)

From (Eq. 1), we find that:

b = 4-3\cdot m

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

-2\cdot m +4-3\cdot m = 2

-5\cdot m = -2

m = \frac{2}{5}

And from (Eq. 1) we find that the y-Intercept is:

b=4-3\cdot \left(\frac{2}{5} \right)

b = 4-\frac{6}{5}

b = \frac{14}{5}

The equation of the line in slope-intercept form is y = \frac{2}{5}\cdot x +\frac{14}{5}.

Then, we obtain the standard form by algebraic handling:

-\frac{2}{5}\cdot x +y = \frac{14}{5}

-2\cdot x +5\cdot y = 14

The equation of the line in standard form is -2\cdot x +5\cdot y = 14.

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: (m = 3, b = 4)

y = 3\cdot x +4

The equation of the line in slope-intercept form is y = 3\cdot x +4.

Then, we obtain the standard form by algebraic handling:

-3\cdot x +y = 4

The equation of the line in standard form is -3\cdot x +y = 4.

4) We begin with a system of linear equations based on the slope-intercept form: (x_{1} = -3, y_{1} = 0, x_{2} = 0, y_{2} = 6)

-3\cdot m + b = 0 (Eq. 3)

b = 6 (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

-3\cdot m+6 = 0

3\cdot m = 6

m = 2

The equation of the line in slope-intercept form is y = 2\cdot x + 6.

Then, we obtain the standard form by algebraic handling:

-2\cdot x +y = 6

The equation of the line in standard form is -2\cdot x +y = 6.

5) We begin with a system of linear equations based on the slope-intercept form: (x_{1} = -1, y_{1} = -2, x_{2} = 5, y_{2} = 3)

-m+b = -2 (Eq. 5)

5\cdot m +b = 3 (Eq. 6)

From (Eq. 5), we find that:

b = -2+m

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

5\cdot m -2+m = 3

6\cdot m = 5

m = \frac{5}{6}

And from (Eq. 5) we find that the y-Intercept is:

b = -2+\frac{5}{6}

b = -\frac{7}{6}

The equation of the line in slope-intercept form is y = \frac{5}{6}\cdot x -\frac{7}{6}.

Then, we obtain the standard form by algebraic handling:

-\frac{5}{6}\cdot x +y =-\frac{7}{6}

-5\cdot x + 6\cdot y = -7

The equation of the line in standard from is -5\cdot x + 6\cdot y = -7.

6 0
2 years ago
Charles is walking arounda fountain that has a circumference of 124ft .what is the diameter of the fountain?
Dovator [93]
Hello.

To find the diameter you use this equation:
Radias<span>cm x 2/pie
</span>
So it would be 124=π*d
d=124/π

So the answer would be :
d=39.47=39.4 feet

Have a nice day
6 0
3 years ago
Read 2 more answers
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Alenkinab [10]

Answer:

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8 0
3 years ago
when a number is added to 1/5 of itself, the result is 24. the equation that models this problem is n 1/5 n = 24. what is the va
GuDViN [60]

Let n be unknown number, then 1/5 of this number is 1/5n. When you add these two numbers, you get

n+\dfrac{1}{5} n.

You know that this sum is equal to 24, so

n+\dfrac{1}{5} n=24.

1. Multiply this equation by 5:

5n+n=120.

2. Solve it:

6n=120,\\n=120:6,\\n=20.

Answer: n=20, correct choice is B.

4 0
3 years ago
Read 2 more answers
Using the method of completing the square, put each circle into the form
tatiyna

Answer:

Standard form: (x-\frac{1}{2})^2 + (y-0)^2 = 15

Center: (\frac{1}{2}, 0)

Radius: r =\sqrt{15}

Step-by-step explanation:

The equation of a circle in the standard form is

(x-h)^{2} + (y-k)^{2} = r^{2}

Where the point (h, k) is the center of the circle

To transform this equation 4x^{2} -4x + 4y^{2} - 59 = 0 this equation  in the standard form we use the method of square.

First, we group similar variables

(4x^{2} -4x) + (4y^{2}) - 59 = 0

Divide both sides of equality by 4

(x^{2} -x) + (y^{2}) - 14.75 = 0

Now we complete square for variable x.

Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:

b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2=  (-\frac{1}{2})^2 = \frac{1}{4}

Now add (\frac{b}{2})^2 on both sides of the equality

(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})

Factor the expression and simplify the independent terms

(x-\frac{1}{2})^2 + (y^{2}) = 15

(x-\frac{1}{2})^2 + (y-0)^2 = 15

Then

h =\frac{1}{2}\\\\k=0

and the center is (\frac{1}{2}, 0)

radius r =\sqrt{15}

3 0
3 years ago
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