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2 years ago

Let a and b be real numbers where a ≠ b ≠ 0. Which of the following functions could represent the graph below?

Mathematics

2 answers:

Ivenika [448]2 years ago

7 0

Answer:

I know this was posted 2 weeks ago, but if it helps, the correct choice is option 2.

Westkost [7]2 years ago

5 0

Answer:

Step-by-step explanation:

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Does anybody know the answers to these? will mark brainliest!

jeyben [28]

Answer:

Tan 45=√2/n

n=√2

again

sin 45=√2/m

m=2

answer: m=2,n=√2

2.

sin 45=x/3

x=3/√2=3√2/2

y=3√2/2

ans:x=3√2/2,y=3√2/3

3.

sin 45=a/4

a=4/√2

b=4/√2

ans:a=4/√2,b=4/√2

b=4 base sides of isosceles triangle

sin 45=4/a

a=4√2

ans:a=4√2 and b=4

5 0

3 years ago

The length of a rectangle is 10 dm more than its width and its perimeter is 80 dm. find its dimensions.

cluponka [151]

Answer:

the width is 15 and length is 25

Step-by-step explanation:

4 0

3 years ago

Please answer this if i get ur wrong my math teacher will give me a detention

liubo4ka [24]

Answer:

No lol

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

What is this expression in factored form? x^2+13x+42

Allushta [10]

Answer and explanation:

$x^{2} + 13x + 42$

(x + 6)(x + 7)

List the multiples of 42

42: 1 2 3 6 7 14 21 42

Then see which of the multiples add up to 13

1 x 42 = 42 --> 1 + 42 $\neq$ 13

2 x 21 = 42 --> 2 + 21 $\neq$ 13

3 x 14 = 42 --> 3 + 14 $\neq$ 13

6 x 7 = 42 --> 6 + 7 = 13

4 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