All categories

3 years ago

Pls help will mark brainlist if correct !! :))

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

E. 31

Step-by-step explanation:

Add the values together.

4 (2/6) + 3 (3/6) + 2 (3/6) = 9 (8/6) or 10 (1/3)

multiply 10 by 3 to find how many thirds there are and add 1 to that number for the extra 1/3.

You might be interested in

Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals intersect at point P. Side QR= 5 m and diagonal QS

svet-max [94.6K]

Answer:

Given the statement: Kite QRST has a short diagonal of QS and a long diagonal of RT. The diagonals intersect at point P.

Properties of Kite:

The diagonals are perpendicular

Two disjoint pairs of consecutive sides are congruent by definition of kite

One diagonal is the perpendicular bisector to the other diagonal.

It is given that: Side QR = 5m and diagonal QS = 6m.

Then, by properties of kite:

$QP = \frac{1}{2}QS$

Substitute the value of QS we get QP;

$QP = \frac{1}{2}(6)$ = 3 m

Now, in right angle $\triangle RPQ$

Using Pythagoras theorem:

$QR^2= RP^2 +QP^2$

Substitute the given values we get;

$(5)^2= RP^2 +(3)^2$

$25= RP^2 +9$

Subtract 9 from both sides we get;

$16= RP^2$

Simplify:

$RP = \sqrt{16} = 4 m$

Therefore, the length of segment RP is, 4m

5 0

3 years ago

Read 2 more answers

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each nig

umka2103 [35]

Answer:no

Step-by-step explanation:

jolu frcu

3 0

4 years ago

How can u tell <br> Whether 1080 is cube number

dolphi86 [110]

TSplit into prime factors

1080 = 2*2*2*3*3*3*5

Its not a cube because although there are 2 triplicates there is also a 5.

Note 2*2*2*3*3*3 is a cube.

3 0

3 years ago

Read 2 more answers

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.