Answer is -66 but show the work of how you got that answer

Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$

3 0

3 years ago

The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?

svet-max [94.6K]

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

$V(t) = V(0)(1-r)^{t}$

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that $r = 0.16$

So

$V(t) = V(0)(1-r)^{t}$

$V(t) = V(0)(1-0.16)^{t}$

$V(t) = V(0)(0.84)^{t}$

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

$V(t) = V(0)(0.84)^{t}$

$0.5V(0) = V(0)(0.84)^{t}$

$(0.84)^{t} = 0.5$

$\log{(0.84)^{t}} = \log{0.5}$

$t\log{0.84} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.84}}$

$t = 3.98$

The half life of the car is 3.98 years.

7 0

3 years ago

For the month of November in a certain city, 88% of the days are cloudy. Also in the month of November in the same city, 54%

Gnoma [55]

Answer:

There is a 61.36% probability that a randomly selected day in November will be foggy if it is cloudy.

Step-by-step explanation:

We have these following probabilities:

An 88% probability that the day is cloudy.

An 54% probability that the day is both foggy and cloudy.

What is the probability that a randomly selected day in November will be foggy if it is cloudy?

This is the percentage of days that are cloudy and foggy divided by those that are cloudy. So:

$P = \frac{0.54}{0.88} = 0.6136$

There is a 61.36% probability that a randomly selected day in November will be foggy if it is cloudy.

4 0

3 years ago

Which type of arc lies outside of the central angle and measures more than 180

Temka [501]

The answer is a major arc.

3 0

3 years ago

Read 2 more answers

Simplify: StartRoot StartFraction 576 Over 64 EndFraction EndRoot The prime factorization of 576 is. The prime factorization of

riadik2000 [5.3K]

The prime factor of the 576 is $2^6 * 3^2$ and the prime factor of the 64 is $2^6$ . And the expression $\sqrt{\dfrac{576}{64}}$ is equal to 3.