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3 years ago

12

Helps please this is due soon

1 answer:

Alinara [238K]3 years ago

4 0

4 not sure how to explain tho

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What is the range of the following function

AleksAgata [21]

The range is the set of y.

(-4, 4) → 4

(-3, 0) → 0

(-2, 1) → 1

(-1, 3) → 3

(0, 0) → 0

(1, 5) → 5

Answer: {0, 1, 3, 4, 5}

3 0

3 years ago

A rectangle's width is 5 feet less than its height. If the area of the rectangle is 564 square feet, what is its length, in feet

larisa86 [58]

We can set up an equation: x * (x + 5) = 564, where x is the length of the rectangle we are trying to solve. Using the quadratic formula to solve this, we eventually come up with approximately 21.38 as a value for x. We can verify this by typing 21.38 * (21.38 + 5) into a calculator and seeing that it approximately equals 564.

5 0

3 years ago

A line passes through the point (9. -9) and has a slope of -4/3.

professor190 [17]

Answer:

$y = \frac{ - 4}{3} x + 3$

Step-by-step explanation:

$using \: point \: slope \: form \\ y - y1 = m(x - x1) \\ y + 9= \frac{ - 4}{3} (x - 9) \\ y + 9= \frac{ - 4}{3}x + 12\\ y = \frac{ - 4}{3} x + 3$

6 0

3 years ago

The formula y - y1 = m(x - x1) is the point-slope form of the equation of a line where m is the slope of the line and (x, y) and

a_sh-v [17]

Answer:

The slope of a line that includes the points (4, -2) and (5, 0) is 2

Step-by-step explanation:

You know that the formula y - y1 = m(x - x1) is the point-slope form of the equation of a line where m is the slope of a line.

The line must include the points (4, -2) and (5, 0). So, being:

(x,y)= (4,-2)
(x1,y1)= (5, 0)

and replacing in the point-slope form of the equation of a line:

-2-0=m(4-5)

You solve the equation for m and get:

-2=m*(-1)

$m=\frac{-2}{-1}$

m=2

<u><em>The slope of a line that includes the points (4, -2) and (5, 0) is 2</em></u>

5 0

3 years ago

5. Find the discriminant of 15x2 = 4x – 1 and describe

diamong [38]

Answer:

a) The discriminant of the equation = - 44

b)The nature of the roots will be imaginary.

c) $x = \frac{2 +\sqrt{11} i}{15} or, x = \frac{2 - \sqrt{11} i}{15}$

Step-by-step explanation:

Here, the given expression is $15x^{2} = 4x -1$

or, $15x^{2} - 4x + 1 = 0$

Now the discriminant (D) of a quadratic equation $ax^{2} +b x + c = 0$

D = $b^{2} - 4ac = (-4)^{2} - 4(15) (1) = 16 - (60) = -44$

Hence, the discriminant of the equation = - 44

As D< 0, so the roots will be imaginary.

Now,by quadratic formula : $x = \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

So, here $x = \frac{-(-4) \pm \sqrt{D} }{2a} = \frac{4 \pm \sqrt{(-44 )} }{30}$

So, either $x = \frac{4 + \sqrt{(-44 )} }{30} or, x = \frac{4 - \sqrt{(-44 )} }{30}$

or, $x = \frac{2 +\sqrt{11} i}{15} or, x = \frac{2 - \sqrt{11} i}{15}$

8 0

3 years ago