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4 years ago

The graph of a linear system is shown. What is the solution to this linear system.

Mathematics

1 answer:

vlabodo [156]4 years ago

4 0

Answer:

(-1,7)

Step-by-step explanation:

The solution is where both lines meet.

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Helppp ...............

vaieri [72.5K]

Answer:

i think its B

Step-by-step explanation:

im not sure but im pretty sure if its wrong im sorry but im pretty sure its correct

5 0

3 years ago

Please any help is necessary

Ugo [173]

Answer:

Option D. 50 ft

Step-by-step explanation:

we know that

The area of the figure is equal to the area of three rectangles

$A=18x+(48-36)(32-11)+(18)(32)$

$A=18x+(12)(21)+(18)(32)\\A=18x+828$

Remember that the area is given

$A= 1,728\ ft^2$

$18x+828=1,728$

solve for x

$18x=1,728-828\\18x=900\\x=50\ ft$

3 0

3 years ago

If the function g(x) = f(x + 2), how can the graph of the function g(x) be obtained from the graph of f(x)?

Yakvenalex [24]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

66 cm

galina1969 [7]

Diameter so i’m assuming it’s one :)

5 0

3 years ago

❊ Simplify :

DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

$\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}$

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

$\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}$

Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:

$\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}$

Add the fractions:

$\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}$

Factor:

$\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}$

Simplify:

$\displaystyle =\frac{a+4}{(a-1)(a+1)}$

We can expand. Therefore:

$\displaystyle =\frac{a+4}{a^2-1}$

Problem 2)

We want to simplify:

$\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}$

Again, let's create a common denominator. First, let's factor out a negative from the second term:

$\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}$

Now to create a common denominator, we can multiply the first term by (a - c) and the second term by (a - b). Hence:

$\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}$

Subtract the fractions:

$\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}$

Distribute and simplify:

$\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}$

Cancel. Hence:

$\displaystyle =\frac{1}{(a-b)(a-c)}$

4 0

3 years ago