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Artyom0805 [142]
3 years ago
13

Solve the equation for P: 3k=7Q+6p

Mathematics
2 answers:
iren [92.7K]3 years ago
7 0

3k=7Q+6p

Solve for P which means we have to make 'p' alone

3k=7Q+6p\\\\ Add  7Q on both sides \\\\ 3k - 7Q = 7Q - 7Q + 6p\\\\ \left ( 3k - 7Q \right ) = 6p\\\\ Divide both side by 6 to make p alone \\\\ \frac{\left ( 3k - 7Q  \right )}{6} = \frac{6p}{6}\\\\ p = \frac{\left ( 3k - 7Q  \right )}{6}

rodikova [14]3 years ago
6 0
P= (3k-7q)/6. That’s all you have to do
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Prove the identity (sina-cosatanb)/(cosa+sinatanb) = tan (a+b)
OLga [1]
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3 years ago
Multiply using the distributive property 12(5y+4).
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Answer:

12(5y+4)

12(9y)

108y

hope it helps and I hope I did it right

Step-by-step explanation:

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3 years ago
Exhibit 6-2 the weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 poun
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3 years ago
Use the distance formula to find the distance between each PowerPoint round to the nearest 10th if necessary
natita [175]

Answer:

Step-by-step explanation:

Distance between two points (x_1,y_1) and (x_2,y_2) is given by the formula,

Distance = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

a). A(-6, -4), B(-3, -1)

Distance between A and B = \sqrt{(-6+3)^2+(-4+1)^2}

                                             = \sqrt{9+9}

                                             = 3√2

                                             = 4.2 units

b). C(3.5, 1), D(-4, 2.5)

Distance between C and D = \sqrt{(3.5+4)^2+(1-2.5)^2}

                                              = \sqrt{58.5}

                                              = 7.65

                                              ≈ 7.7 units

c). Distance between X(5, -5) and Y(-5, 5)

    Distance = \sqrt{(5+5)^2+(-5-5)^2}

                    = \sqrt{200}

                    = 10√2

                    = 14.14

                    ≈ 14.1 units

4 0
3 years ago
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