Write two decimals that are equivalent to the given decimal. 0.60 and 0.9

Solve the system by elimination 4x-y=2 and x+3y=7

alexdok [17]

Answer:

x=1 y=2

Step-by-step explanation:

Isolate x for 4x -y=2: X=2+y/4

Substitute x= 2+y/4

(2+y/4 )+3y=7

simplify 2+13y/4=7

Isolate y for 2+13y/4=7

y=2

Forx= 2+y/4

Substitute y = 2

x= 2+2/4

Simplify

X=1

3 0

2 years ago

What is the measure of angle x? 50 degrees 55 degrees О 60 degrees 80 degrees

Yanka [14]

Answer:

it may be 115 if its a quadrilateral

3 0

2 years ago

Read 2 more answers

a point on a perpendicular bisector is 7cm from each endpoint of the bisected segment and 5cm from the point of intersection. Wh

love history [14]

9514 1404 393

Answer:

about 9.80 cm

Step-by-step explanation:

The length of half the segment (h) can be found from the Pythagorean theorem:

h² +5² = 7²

h² = 7² -5² = 49 -25 = 24

h = √24 = 2√6

This is half the segment length, so the whole segment length is ...

L = 2h = 2(2√6)

L = 4√6 ≈ 9.7980

The length of the segment is 4√6 ≈ 9.80 cm.

8 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.