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3 years ago

Which equation best matches the graph shown below ?

Mathematics

1 answer:

Shalnov [3]3 years ago

8 0

Pretty sure it’s the bottom right option

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Given the equation 2x +4 = 4x -2, select the reasoning that correctly solves for x. A.add 2, subtract 2x, then divide by 2. B.ad

Anestetic [448]

2 x + 4 = 4 x - 2
2 x + 4 + 2 = 4 x - 2 + 2
2 x + 6 = 4 x
2 x - 2 x + 6 = 4 x - 2 x
6 = 2 x
6 : 2 = 2 x : 2
x = 3
Answer: A ) add 2 , subtract 2 x, then divide by 2

5 0

3 years ago

Read 2 more answers

Question 40 and it about adding and subtracting polynomials

kolbaska11 [484]

Answer:

39x-7

Step-by-step explanation:

The perimeter is all the sides added together so

9x + 8x - 2 + 5x + 1 + 17x - 6

Since you can't solve it you just want to simplify the equation, add the like terms together

9x+8x+5x+17x = 39x

-2+1-6 = -7

Your equation is 39x-7

3 0

4 years ago

What is the answer?<br><br> Y= -x-4 <br> Y=x

Mashcka [7]

Do you mean like this?

Y = -x - 4
Y = x

x = -x - 4
2x = -4
x = -2

Therefore y = -2

7 0

3 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

3n+4 first the 7 terms

kiruha [24]

Answer:

4, 7, 10, 13, 16, 19, 22

Step-by-step explanation:

Not sure if there is more to this question but here is what I assume:

We want to find the first 7 terms of this equation given 3n+4

Assuming we are starting at 0 we plug 0 in for n.

3(0) + 4 → 4

So our first answer is 4

We continue by pugging in the next numbers after that to get to the first 7 terms

3(1) + 4 → 7

3(2) + 4 → 10

3(3) + 4 → 13

3(4) + 4 → 16

3(5) + 4 → 19

3(6) + 4 → 22

5 0

3 years ago