Ask question

Ask question

All categories

3 years ago

12

E/P 14)The equation 2x²+ 4kx - 5k = 0, where k is a constant, has no real roots.

1 answer:

aalyn [17]3 years ago

4 0

Step-by-step explanation:

Discriminant = 16k² - 4(2)(-5k)

Since 16k² + 40k < 0, 2k² + 5k < 0, k(2k + 5) < 0.

Therefore -5/2 < k < 0.

You might be interested in

I NEEED HELP ON THIS PLSSS.

Nuetrik [128]

The answer is f!!!!!!!!!!!!!!!

4 0

3 years ago

Read 2 more answers

What is the inverse of the function f(x)=2x-10?

8090 [49]

Answer: Try D

Step-by-step explanation:

5 0

3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

What's the answer?????????????

Taya2010 [7]

Answer:

BD = 12 :)

Step-by-step explanation:

Alright, we'll need the Pythagorean theorem for this!

So, the length of AC is 10. That means the lengths of AD and DC are both half of that, which is 5 :)

DC = 5

We already know that BC = 13, so we can plug in these values into the pythagorean theorem for the right triangle BDC:

BD^2 + DC^2 = BC^2

BD^2 + 5^2 = 13^2

BD^2 + 25 = 169

BD^2 = 169 - 25 = 144

BD = √144 = 12 :)

3 0

3 years ago

What is 4x+3y+2z and 3x+5y+3z?

My name is Ann [436]

Answer:

Step-by-step explanation:

7x + 8y + 5z

I put the equations in parentheses, assuming that I would be adding both of the equations together. Then, I removed them and combined the like terms. I used two calculators to check my answer, and this should be correct.

6 0

2 years ago

Read 2 more answers