I need to find the value of x?

Solve the system of linear equations by substitution.

SVEN [57.7K]

Answer:

Answer:

x= 29/11 or y= 25/11

Step-by-step explanation:

2x - y= 3 ...........equation 1

x + 5y= 14 ............equation 2

Make x the subject of the formula in equation 2

x= 14 - 5y ..............equation 3

Substitute x=14 - 5y in equation 1

2(14- 5y) - y=3

28 - 10y - y=3

Collect like terms

-10y - y=3 - 28

-11y= -25

divide both sides by coefficient of y

-11y/-11 = -25/-11

y= 25/11

Substitute y= 25/11 in equation 3

x=14 - 5(25/11)

x= 14 - 125/11

x= 29/11

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Ben and Chi are painting a room in the city recreation center. They started with 4 gallons of paint. On the first day, Ben used

Nuetrik [128]

Wouldnt that be zero cuz they used more then they started of with?

6 0

3 years ago

I'm confused on this

noname [10]

Answer:

AB = 33.78228208

BD = 42.3

AT= 20.33068487

Area= 859.9879701

Step-by-step explanation:

Using the sin rule:

a b

If I use A = 90, a= 42.3

B = 53, b= AB

I get AB = 33.78228208 units

Then the diagonal is already 42.3 cm so I think. BD should be 42.3 cm too.

I used the sin rule again for ABD to find AD but by using B as 37° and b as AD.

I get AD = 25.45677548

For ADT,

AD is the hypotenuse.

Now I get from the same sin rule,

A= 90°. a= 25.45677548

B= 53°. b = AT

AT= 20.33068487 units.

Area of rectangle = l×b

= AB × AD

so that gives me 859.9879701 square units

7 0

3 years ago

Please help me with this question please is so hard

den301095 [7]

Answer:

1) $101,400

2) $27,000

Step-by-step explanation:

1) For this, we can first find how much she makes in a year from her salary alone:

$2,200×12= $26400.

Now, she also received a commission of 2.5%, so we can take this percentage from 3 million to find the total from commission:

3,000,000×0.025= $75,000

Adding these two gives us a total of $101400.

2) We can use the Simple Interest Formula to solve this (I= Prt). Plug in the values given to give us:

I= 5000(1.08)(5)

I= $27000

4 0

4 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago