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3 years ago

Quadratic functions heeeeeellllppp

Mathematics

1 answer:

anzhelika [568]3 years ago

6 0

Answer:

See below

Step-by-step explanation:

Remember that quadratic functions are parabolas when graphed. The solutions are where the parabola crosses the x-axis.

1. The vertex of the parabola in f(x) is (0, 9) which is above the x-axis and the parabola opens up. So the parabola does not cross the x-axis. Therefore the solutions are imaginary.

2. The vertex of the parabola in g(x) is (9, 0) which is on the x-axis and parabola opens up. Therefore, there is a double solution.

3. The vertex of the parabola in h(x) is (-1, -9) which is below the x-axis and the parabola opens up. Therefore, there are two real solutions.

I know this is a long explanation, but that is a way of looking at the problem.

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7.00 and 7.40

Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

PLEASE HELP VERY URGENT

gavmur [86]

Answer:

B) $(2, 1)$

Step-by-step explanation:

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6 0

3 years ago

Area and percent .

siniylev [52]

Answer:

a. 61.92 in²

b. 21.396 ≈ 21.4%

c. $4.71

Step-by-step explanation:

a. Amount of waste = area of rectangular piece of stock - area of two identical circles cut out

Area of rectangular piece of stock = 24 in × 12 in = 288 in²

Area of the two circles = 2(πr²)

Use 3.14 as π

radius = ½*12 = 6

Area of two circles = 2(3.14*6²) = 226.08 in²

Amount of waste = 288 - 226.08 = 61.92 in²

b. % of the original stock wasted = amount of waste ÷ original stock × 100

= 61.92/288 × 100 = 6,162/288 = 21.396 ≈ 21.4%

c. 288 in² of the piece of stock costs $12.00,

Each cut-out circle of 113.04 in² (226.08/2) will cost = (12*113.04)/288

= 1,356.48/288 = $4.71.

3 0

4 years ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

3 years ago