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jasenka [17]
3 years ago
12

I need a models or visuals for that problem​

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0

Answer:I will be a mode

Step-by-step explanation:

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Find the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)
Eva8 [605]

The coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4) is (\frac{-17}{5} ,\frac{2}{5} )

Explanation :

the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Apply section formula to find coordinates of point Q

(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} )

Ratio m:n is 2:3 and point R is (x1,y1) , point S is (x2,y2)

Substitute all the values inside the formula

(\frac{2(2)+3(-7)}{2+3} , \frac{2(4)+3(-2)}{2+3} )\\\\(\frac{-17}{5} ,\frac{2}{5} )

The coordinates of point Q is (\frac{-17}{5} ,\frac{2}{5} )

Learn more :  brainly.com/question/13878373

7 0
3 years ago
A farmer plants apple, pear and cherry trees
gogolik [260]
? is there any more context to the question?
8 0
3 years ago
An item was sold at a loss of 20%. Had it been sold for Rs 3000more then the profit would have been 10%. What is the cost price
vagabundo [1.1K]

Answer:

Cost Price = Rs 10000

Step-by-step explanation:

Assume:

Cost of the item = x

Item was sold at a loss of 20%:

Loss = 20% of x = 0.2x

Item sold = x - 0.2x = 0.8x

Item sold at a profit of 10%:

Profit = 10% of x = 0.1x

item sold = x + 0.1x = 1.1x

Solve:

Difference = 1.1x - 0.8x = 0.3x

0.3x = Rs 3000

x = Rs 3000 ÷ 0.3

x = Rs 10000

3 0
3 years ago
In △ABC, m∠A=39°, a=11, and b=13. Find c to the nearest tenth.
Talja [164]

For this problem, we are going to use the <em>law of sines</em>, which states:

\dfrac{\sin{A}}{a} = \dfrac{\sin{B}}{b} = \dfrac{\sin{C}}{c}


In this case, we have an angle and two sides, and we are trying to look for the third side. First, we have to find the angle which corresponds with the second side, B. Then, we can find the third side. Using the law of sines, we can find:

\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{B}}{13}


We can use this to solve for B:

13 \cdot \dfrac{\sin{39^{\circ}}}{11} = \sin{B}

B = \sin^{-1}{\Big(13 \cdot \dfrac{\sin{39^{\circ}}}{11}\Big)} \approx 48.1


Now, we can find C:

C = 180^{\circ} - 48.1^{\circ} - 39^{\circ} = 92.9^{\circ}


Using this, we can find c:

\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{92.9^{\circ}}}{c}

c = \dfrac{11\sin{92.9^{\circ}}}{\sin{39^{\circ}}} \approx \boxed{17.5}


c is approximately 17.5.

8 0
3 years ago
An archer has a mean distance of 1.6 inches and a MAD distance of 1.3 inches in the first round. In the second round, the archer
schepotkina [342]
I think it’s 1.5 cause it shows it
7 0
3 years ago
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