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Naily [24]
3 years ago
7

What two numbers multiply to -84 but add to 5

Mathematics
2 answers:
Norma-Jean [14]3 years ago
7 0
-7 * 12 = -84
       but
-7 + 12 = 5
Rudik [331]3 years ago
7 0
7 times -12 equals -84.

7 + -12 equals 5.

The answer is 7 and -12.

Hope this helped!
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-4(w+1)=-24<br><br> what does w equal?
makvit [3.9K]

Answer:

w=5

Step-by-step explanation:

using the distributive property to simplify the equation

-4(w)+1(-4)=-24 so

-4w-4=-24 (adding 4 to both sides)

-4w==20 (divide by -4)

w=5

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You begin the week with 20 units of milk. You purchase 40 units. Your ending inventory for the week is 30 units. How many units
Musya8 [376]

Answer:

30 units.  

Step-by-step explanation:

Let x be number of milk units sold.

It has been given that You begin the week with 20 units of milk. You purchase 40 units. Therefore, total units of milk will be 20+40=60 units.

It has been given that your ending inventory for the week is 30 units, which means after selling x units from 60 units you are left with 30 units.

We can set this information in an equation as:

60-x=30

-x=30-60

-x=-30

x=30

Therefore, you have sold 30 units.

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.

((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.

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Slav-nsk [51]
About 1.204972095x10^20 joules in a year
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