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Find the solution to the system of equations 2x+3y+z=20 3y+5z=12 y-5z=4

podryga [215]

Answer:

x =4 ,y =4 and z =0 is the solution.

Step-by-step explanation:

We shall solve the equations using elimination method

In equation 2 and 3 we can see that coefficient of z in both equations are same with opposite sides , adding equations (2) and (3)

3y+5z = 12

y -5z = 4 adding the equations

___________

4y = 16

dividing both sides by 4

y = $\frac{16}{4}$

y = 4

Plugging the value of y in equation 2 or equation 3 ,we get

3(4) +5z = 12

12+5z = 12

5z = 12-12 or 5z =0 gives z =0

plugging y =4 and z =0 in equation 1

2x+3(4)+ 0 = 20

2x+12 = 20

2x =20-12

2x = 8

x = 8 divided by 2 gives

x =4

therefore solution of the system is given by

x=4 , y =4 and z =0

3 0

3 years ago

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The vertices of a right triangle are (0, 0), (1, 0), and (0, 1).

Alchen [17]

Answer:

(3,3)

Step-by-step explanation:

Given

$Vertices: (0,0)\ (1,0)\ (0,1)$

$Scale\ Factor = 3$

Required

Determine which can't be any of the new vertices

First, we need to determine the new vertices:

$New\ Vertex = Scale\ Factor * Old\ Vertex$

For (0,0):

$New\ Vertex = 3 * (0,0)$

$New\ Vertex = (3 * 0,3 * 0)$

$New\ Vertex = (0,0)$

For (1,0):

$New\ Vertex = 3 * (1,0)$

$New\ Vertex = (3 * 1,3 * 0)$

$New\ Vertex = (3,0)$

For (0,1):

$New\ Vertex = 3 * (0,1)$

$New\ Vertex = (3 * 0,3 * 1)$

$New\ Vertex = (0,3)$

Comparing the calculated new vertices to the list of given options; (3,3) can't be any of the new vertices of the new triangle

5 0

3 years ago

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

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F(x)=-2X-3 and g(x)=5X-6. Find 2f(x)-3g(x)+F(x)

MakcuM [25]

Answer:

-21X + 9

Step-by-step explanation:

8 0

3 years ago

Kyra see a pair of boots that costs $60.00.They are on salenfor 25% off.She also has a coupon for an additional 10% off.How much

nikitadnepr [17]