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1 year ago

What are the solutions to this quadratic equation x2+6x-5=0

Mathematics

2 answers:

marishachu [46]1 year ago

8 0

Answer:

See Below.

Explanation:

Quadratic Form:

$a {x}^{2} + bx + c$

Quadratic Formula:

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

Based on the information given from the question, we can deduce:

a = 1

b = 6

c = -5

Now we can substitute all these values into the formula to find x.

$x = \frac{ - 6 + - \sqrt{ {6}^{2} - 4(1)( - 5)} }{2(1)} \\ = - 3 + \sqrt{14} \: or \: - 3 - \sqrt{14}$

ale4655 [162]1 year ago

6 0

Thus, the solutions of the quadratic equation x 2 + 5x + 6 = 0 are x = -2 and x = -3.

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A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?

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Hello,

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HELP Sam measures the height of 7 different scoops of ice cream. The mean of the heights was 3.8 cm. Which of the following best

OLga [1]

Sam measures the height of 7 different scoops of ice cream. The mean height was 3.8 cm. If all the scoops are evenly distributed, they would all be 3.8 cm tall.

<h3>How to find the mean value of a data set?</h3>

The mean value of the data set is the ratio of the sum of the data set's values to the number of values it has.

If the data set consists of values

x_1, x_2, ..., x_n

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$\overline{x} = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$

Sam measures the height of 7 different scoops of ice cream. The mean height was 3.8 cm.

$\overline{x} = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$

mean= $\dfrac{(a+b+c+d+e+f+g)}{7}$

Therefore, If all the scoops are evenly distributed, they would all be 3.8 cm tall.

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Which inverse operation would be use to verify the following 102 divided by 3= 34

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3 years ago

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A

julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$ (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$ (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$