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2 years ago

What is the solution of the system of equations shown in the graph ?

Mathematics

1 answer:

Katarina [22]2 years ago

4 0

The answer is (0,-3) and (0-1)

hope this helped

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A triangle has one side length of 12 inches and another of 8 inches. Describe all possible lengths of the third side. Show and e

katrin [286]

Answer:

The third side must be smaller than 20 inches and greater than 4 inches

$4$

Step-by-step explanation:

Let a, b and c be the lengths of triangle's sides. Then

$a+b>c\\ \\a+c>b\\ \\b+c>a$

Use this rule in your case. So, if a=12 and b=8, then

$12+8>c\\ \\12+c>8\\ \\8+c>12$

Hence, you get

$c-4\\ \\c>4$

From these inequalities, you can state that

$4$

So, c must be smaller than 20 inches and greater than 4 inches.

5 0

3 years ago

Read 2 more answers

Give the correct name for<br> 2x3 + 5x

NikAS [45]

Answer: x(2x²+5)

please give brainliest if it helped

8 0

3 years ago

The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

3 years ago

To play basketball with her friends, Evangeline needs to pump air in her ball, which is completely deflated. Before inflating it

Nadya [2.5K]

Answer:

ρ_air = 0.15544 kg/m^3

Step-by-step explanation:

Solution:-

- The deflated ball ( no air ) initially weighs:

m1 = 0.615 kg

- The air is pumped into the ball and weight again. The new reading of the ball's weight is:

m2 = 0.624 kg

- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.

m_air = Δm = m2 - m1

m_air = 0.624 - 0.615

m_air = 0.009 kg

- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:

V_air = 4*π*r^3 / 3

V_air = 4*π*0.24^3 / 3

V_air = 0.05790 m^3

- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:

ρ_air = m_air / V_air

ρ_air = 0.009 / 0.05790

Answer: ρ_air = 0.15544 kg/m^3

4 0

3 years ago

Read 2 more answers

Help me pleaseeeee :(

Sergeu [11.5K]

Answer:

-4

Step-by-step explanation:

7 0

3 years ago