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3 years ago

What are the terms in the expression 6x + 5y + 3

Mathematics

2 answers:

Luba_88 [7]3 years ago

5 0

6x,5y and 3 those are terms

Katen [24]3 years ago

4 0

The terms should be 6x and 5y

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2 is 4% of what number?

kaheart [24]

$x-searched\ number\\\\4\%x=2\\\\0.04x=2\ \ \ \ |:0.04\\\\x=50\\\\2\ is\ 4\%\ of\ 50$

6 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

2 years ago

Can I please get some help with this :D

Luba_88 [7]

1. Anwser=53.3 (do 40*3 then anwser x 4)
2anwser=120

6 0

2 years ago

The hypotenuse of a 30°-60°-90°

kramer

asdfghjklqwertyuiohcjbahckdsahcajdnc

3 0

2 years ago

Find all angles and pls show steps

Ghella [55]

Answer:

Both angles have a measure of 134degrees, y = 27degrees.

Step-by-step explanation:

As per what is given in the problem:

There are 2 parallel lines, both are intersected by a transversal.

Remember the theorem, when two parallel lines are intersected by a transversal, then the alternate exterior angles are congruent.

The is meanse that:

3y + 53 = 7y - 55

Solve using inverse operations:

3y + 53 = 7y - 55

+55 +55

3y + 108 = 7y

-3y -3y

108 = 4y

/4 /4

27 = y

Now, substitute back in to find the value of the angle:

3y + 53

y = 27

3 ( 27 ) + 53

81 + 53

= 134

Since the angles are alternate exterior, they are congruent, hence both angles have a measure of 134degrees.

7 0

2 years ago