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3 years ago

8

50 POINTS for both questions

2 answers:

vitfil [10]3 years ago

7 0

Answer:

thats 5 points why u sayin 50

Step-by-step explanation:

love history [14]3 years ago

6 0

Answer:

3. i think 15.5

4. 140*

Step-by-step explanation:

in number 4, d is a vertical angle to the angle measured at 140.

in number 3, QV=SV, so 17=4x-1 18=4x 18/4=x x=4.5

and TS=TQ, so 3(x) + 2 = 3(4.5) + 2 = 13.5 + 2 = 15.5 =TQ

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HELP ASAP <br><br> Find the zeros of y = x^2 - 4x - 9 by completing the square.

Alla [95]

Answer: x= 2 + √ 13 ,2- √ 13

Step-by-step explanation:

Replace y with 0 and solve for x

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3 years ago

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Ethan has 3/4 of a box of cereal. He gives 1/8 of the box to his little brother and Ethan eats 1/4 of the box, himself. What fra

riadik2000 [5.3K]

Answer:

<em>3/8 of the box cereal</em>

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3 years ago

A watch loses 15 seconds every day how many minutes will it lose in 2 weeks.<br>

REY [17]

2 weeks = 14 day

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2 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

3 years ago

3x2 – 7x 7x = 4x – 8

ratelena [41]

Answer:

x =−1−√93/23,−1+√93/23

Step-by-step explanation:

x=0.37581090…,−0.46276742...

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3 years ago

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