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yaroslaw [1]

The solutions to f(x) = 64 is x = 7 and x = –7.

Solution:

Given data:

$$f(x)=x^{2}+15$ – – – – (1)

$f(x)=64$ – – – – (2)

To find the solutions to f(x) = 64.

Equate equation (1) and (2), we get

$x^2+15=64$

Subtract 15 from both sides of the equation.

$x^2+15-15=64-15$

$x^2=49$

$x^2=7^2$

Taking square root on both sides of the equation, we get

x = ±7

The solutions to f(x) = 64 is x = 7 and x = –7.

5 0

3 years ago

The coordinates of the endpoints of ST are (-2,3) and (3, y) respectively supposethe distance between s and t us 13 units.What v

igor_vitrenko [27]

Answer:

The values of y would be -9 and 15

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$d=13\ units$

S(-2,3) and T(3,y)

substitute the given values in the formula and solve for y

$13=\sqrt{(y-3)^{2}+(3+2)^{2}}$

$13=\sqrt{(y-3)^{2}+25}$

squared both sides

$169=(y-3)^{2}+25$

$(y-3)^{2}=169-25$

$(y-3)^{2}=144$

take square root both sides

$y-3=(+/-)12$

$y=3(+/-)12$

$y=3(+)12=15$

$y=3(-)12=-9$

therefore

The values of y would be -9 and 15

5 0

4 years ago

A.

Reptile [31]

Answer:

The real zeros of f(x) are x = 0.3 and x = -3.3.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem, we have that:

$f(x) = x^{2} + 3x - 1$

So

$a = 1, b = 3, c = -1$

$\bigtriangleup = 3^{2} - 4*1*(-1) = 13$

$x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3$

$x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3$

The real zeros of f(x) are x = 0.3 and x = -3.3.

7 0

3 years ago

Find the value of x that would prove this is an isosceles trapezoid. Then find each segment length.

vredina [299]

Answer:

Check pdf

Step-by-step explanation:

Download pdf

4 0

3 years ago

Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows

never [62]

Answer:

a. For the first lake;

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m·t/200 - m·t²/80000

b. t ≈ 1.00505 hours

c. 200 hours

Step-by-step explanation:

The flow rate of water in and out of the lakes = 500 liters/hour

The volume of water in the first lake = 100 thousand liters

The volume of water in the second lake = 200 thousand liters

The mass of toxic substances that entered into the first lake = 500 kg

The concentration of toxic substance in the first lake = m₁/(100000)

Therefore, we have;

The quantity of fresh water supplied at t hours = 500 × t

The change

The change in the mass of the toxic substance with time is given as follows

dm/dt = (m - m/100000 × 500 × t)/100000

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m/100000 × 500 × t - (m/100000 × 500 × t)/200000 × 500 × t

Using an online tool, we have;

c = m·t/200 - m·t²/80000

b. When c < 0.001 kg per liter, we have m < 0.001 × 100000, which gives m < 100

Substituting gives;

0.001 = (100·t - 0.005·100·t²)/100000, solving with an online tool, gives;

t ≈ 1.00505 hours

c. For maximum concentration, we have;

c = m·t/200 - m·t²/80000

m/200000 = m·t/200 - m·t²/80000

1/200000 = t/200 - t²/80000

dc/dt = d(t/200 - t²/80000)/dt = 0

Solving with an online tool gives t = 200 hours

6 0

3 years ago