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kaheart [24]
3 years ago
14

Please help Help Help

Mathematics
1 answer:
larisa [96]3 years ago
6 0

Answer:

6x-4y=24

Step-by-step explanation:

From the graph, there are two intersections to be found one at (4,0) and another at (0,-6)

Now,

slope,m= y1-y2/x1-x2

=0+6/4-0

=3/2

The equation of the straight line passingt through the point (4,0) is

y-y1=m(x-x1)

or, y-0=3/2 (x-4)

or, 2y=3x-12

or, 3x-2y=12

or, 2(3x-2y)=2(12) [Multiplying both sides by 2]

or, 6x-4y=24

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ed has 8 pieces of candy which represents 40% of all the candy in his home, how many pieces are in his home
I am Lyosha [343]
C = total number of candy
.4(c) = 8 pieces of candy
Divide c by .4: c= 20
20-8=12

Ed has 12 pieces of candy at home
7 0
3 years ago
What are the factors of the equation x 2 - 5x + 6?
xxTIMURxx [149]

Answer:

(x-3)(x-2)

Step-by-step explanation:

x^2 -5x+6

What two numbers multiply to 6 and add to -5

-3 * -2 = 6

-3 -2 = -5

(x-3)(x-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
2 years ago
For AB, point A has coordinates (3, -5) and point B has coordinates (-1, 9). What are the coordinates of the midpoint of AB?
kow [346]
The coordinates midpoint of AB are (1,2)
5 0
3 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Troyanec [42]

Answer: 300

Step-by-step explanation:

Let the number be represented by x.

Therefore, based on the information given in the question, this can be expressed as:

36% of x = 108

36/100 × x = 108

0.36 × x = 108

0.36x = 108

x = 108/0.36

x = 300

Therefore, the number is 300

8 0
3 years ago
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