How do you solve this linear systems of equations? x+3y−z=24 x+2y+5z=13 x+z=12

timama [110]

I pretty sure your answer is yes but correct me if i’m wrong

6 0

3 years ago

Trisha helped clean up her neighborhood by picking up plastic she collected 3/4 pound of plastic the first day and 1/6 pound of

AysviL [449]

She collected more 1/3 pounds on the first day

7 0

2 years ago

Find the equation of the line with y-intercept 5 and slope 5/9

konstantin123 [22]

Answer:

y=5/9x+5

Step-by-step explanation:

The formula is y=mx+b with m being the slope and b being the y intercept so just plug in the values.

3 0

2 years ago

Read 2 more answers

Connor is going to see a movie and is taking his 4 kids. Each movie ticket costs $13 and there are an assortment of snacks avail

EleoNora [17]

If Connor buys 3 snacks at 3 dollars each for himself and his 3 children, the net cost (including movie tickets) would be 88 dollars.

If he buys x amount of snacks, his cost will be determined by C=52+3x or C=52+3*x (both are equivalent, but not everyone knows every way to represent multiplication)

Step-by-step explanation:

To find the cost of the snacks:

Cost = Price*NumberOfSnacks, or C=p*n, or C=pn

To find the unit cost of the snack, find the price, the p value, which is 3 dollars.

Then, find the total number of snacks that will be bought. Since each person receives an equal number of snacks, 3 per person, and there are 4 people (himself and his 3 children), we can multiply 3*4 to find the n value, which is now known to be 12.

C=3*12=36. Add to the 13 dollar per person charge for tickets (52 dollars for 4 people), and you get 88 dollars.

To find the cost if he buys x amount of snacks for himself and his 3 children (assumed to be "everyone"), all costs but the n value are to be the same, creating this equation:

C=52+3x

In this context, 52 is the cost of movie tickets, 3 is the cost of the snacks, and x is the total number of snacks bought. C represents total cost.

5 0

2 years ago

Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137

Nata [24]

Answer:

Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle $\triangle ABC$ with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles $\angle A, \angle B, \angle C$

As per Law of cosine:

$cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}$

Putting the values of a,b and c to find out angles $\angle A, \angle B, \angle C$ .

$cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ$

$cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ$

$cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ$

Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

6 0

3 years ago