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Brums [2.3K]
2 years ago
15

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:

b.

Step-by-step explanation:

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Does anybody know the answers to these? will mark brainliest!
jeyben [28]

Answer:

1.

Tan 45=√2/n

n=√2

again

sin 45=√2/m

m=2

answer:<u> </u><u>m</u><u>=</u><u>2</u><u>,</u><u>n</u><u>=</u><u>√</u><u>2</u>

<u>2</u><u>.</u>

sin 45=x/3

x=3/√2=3√2/2

y=3√2/2

<u>a</u><u>n</u><u>s</u><u>:</u><u>x</u><u>=</u><u>3√2/2</u><u>,</u><u>y</u><u>=</u><u>3√2/</u><u>3</u>

<u>3</u><u>.</u>

sin 45=a/4

a=4/√2

b=4/√2

ans:<u>a</u><u>=</u><u>4</u><u>/</u><u>√</u><u>2</u><u>,</u><u>b</u><u>=</u><u>4</u><u>/</u><u>√</u><u>2</u>

4.

b=4 base sides of isosceles triangle

sin 45=4/a

a=4√2

ans:<u>a</u><u>=</u><u>4</u><u>√</u><u>2</u><u> </u><u>and</u><u> </u><u>b</u><u>=</u><u>4</u>

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:

No

8 0
2 years ago
Read 2 more answers
What is this expression in factored form?<br> 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
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