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IrinaVladis [17]
3 years ago
6

Write the quadratic equation whose roots are 4 and −4 , and whose leading coefficient is 2 .

Mathematics
1 answer:
Vlada [557]3 years ago
7 0
I would love to help you but I’m supper I can f
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The original price is $40.90 and the sale price is 30% off what would be two different expressions for the equation?
Ann [662]

Answer:

(40.90 ÷ 100) × 30

And

0.3 × 40.90

8 0
3 years ago
Confirming my responses are accurate
Rina8888 [55]

Answer:

For the first question, the answer is rotation 290 counterclockwise.

And for the second question, the answer is accurate.

8 0
3 years ago
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.8 parts
Alexxandr [17]

Answer:

The value of the test statistic is t = 2.67

Step-by-step explanation:

The null hypothesis is:

H_{0} = 7.8

The alternate hypotesis is:

H_{1} \neq 7.8

Our test statistic is:

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

In this problem, we have that:

X = 8.2, \mu = 7.8, \sigma = 0.6, n = 16

Then

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

t = \frac{8.2 - 7.8}{\frac{0.6}{\sqrt{16}}}

t = 2.67

The value of the test statistic is t = 2.67

5 0
3 years ago
The graph is of y = f(x) is shown below. What are all of the real solutions of f(x) = 0?
Reika [66]

The real solutions of f(x) = 0 is; x = -8, 0 and 4

<h3>How to find the roots of a polynomial graph?</h3>

When talking about real solutions of a polynomial, we are simply referring to the values of x that make the polynomial f(x) = 0.

Now, in a polynomial graph as attached, the real solutions are the roots and they are the values of x where the curve crosses the x-axis.

From the given graph, the real solutions are at x = -8, 0 and 4

Thus, we conclude that the real solutions of f(x) = 0 is; x = -8, 0 and 4

Read more about Polynomial roots graph at; brainly.com/question/14625910

#SPJ1

8 0
1 year ago
PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
drek231 [11]

Answer:

1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

6 0
3 years ago
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