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2 years ago

What is the solution to the linear system of equations?

1 answer:

3 0

The answer is C

you would isolate x in the first equation then substitute it in to the second equation and solve for y. then plug y in to the first equation.

You might be interested in

How to solve #4 and #5

MArishka [77]

If SU bisects TSV, then TSU = USV
4y + 11 = 6y + 5
6y - 4y = 11 - 5 = 6
y = 6/2 = 3

Therefore, m<TSU = 4(3) + 11 = 12 + 11 = 23

7 0

3 years ago

The table represents a linear function.

Iteru [2.4K]

What are the x values of the table?

3 0

2 years ago

A bowl has a radius of 3 inches. What is the circumference of the bowl?

madreJ [45]

Answer:

9.42 (sorry if im late)

Step-by-step explanation:

I believe that if you take the radius and multiply it by pi (3.14), you get the circumfrence

3.14 x 3 = 9.42

6 0

2 years ago

Find the slope of the line that passes through (8,9) and (5, 14).

AleksAgata [21]

Answer:

6/-4 (right 6 points on x-axis & down 4 points on y- axis

Step-by-step explanation:

14- 8 6

---------- = ------

5-9 - 4

7 0

2 years ago

☆pls help no one has helped me and I'm so confused!

motikmotik

Answer:

Step-by-step explanation:

a). $\frac{5\pm\sqrt{-4}}{3}$ = $\frac{5}{3}\pm \frac{\sqrt{-4}}{3}$

= $\frac{5}{3}\pm\frac{2\sqrt{(-1)} }{3}$

= $\frac{5}{3}\pm\frac{2i}{3}$ [Since, i = $\sqrt{(-1)}$ ]

b). $\frac{10\pm\sqrt{-16}}{2}$ = $\frac{10}{2}\pm \frac{\sqrt{-16}}{2}$

= $5\pm \frac{4\sqrt{-1}}{2}$

= 5 ± 2i [Since, i = $\sqrt{(-1)}$ ]

c). $\frac{-3\pm \sqrt{-144}}{6}$ = $-\frac{3}{6}\pm \frac{\sqrt{-144}}{6}$

= $-\frac{3}{6}\pm \frac{12\sqrt{-1}}{6}$

= $-\frac{1}{2}\pm 2\sqrt{-1}$

= $-\frac{1}{2}\pm 2i$ [Since, i = $\sqrt{(-1)}$ ]

4 0

2 years ago