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1 year ago

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17) Given the point A(-3, -2) and B(6, 1), find the coordinates of the point P on directed line segment AB thatpartition AB in t

he ratio 2:1.

1 answer:

DaniilM [7]1 year ago

6 0

We find the directed line segment AB, as follows:

We subtract the x and y component of the first coordinate from the second one, that is:

$AB=(6-(-3),1-(-2))\Rightarrow AB=(9,3)$

Now, we proceed as follows:

We will use the ratio 2:1 to solve in the following expression to find the x and y coordinates for the point P:

$(x_1+\frac{a}{a+b}(x_2-x_1),y_1+\frac{a}{a+b}(y_2-y_1))$

This expression is for a rate of the form a:b.

That is:

$(-3+\frac{2}{2+1}(6-(-3)),-2+\frac{2}{2+1}(1-(-2)))=(3,0)$

From this, we have that the point p is (3, 0).

***

You can only find the coordinates of the P point by using the formula

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PLEASE HELP 2O POINTS

vovangra [49]

C. The first is a solution, but the second is not.

5x - y/3 = 13 ; (2,-9)
substitute the letters:
5(2) - (-9/3) = 13
10 - (-3) = 13 : note that deducting a number with a negative sign turns both sign as positive.
10 + 3 = 13 ;
13 = 13

5x - y/3 = 13 ; (3,-6)
5(3) - (-6/3) = 13
15 - (-2) = 13
15 + 2 = 13
17 = 13 not equal. not a solution

hope i could help

4 0

3 years ago

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In parallelogram HIJK, the measure of angle H is 45 degrees. 1.Find the measure of J and explain how you know

s2008m [1.1K]

Answer:

45°

Step-by-step explanation:

A parallelogram is a polygon with 4 sides and 4 angles (that is a quadrilateral).

For a parallelogram ABCD, Opposite sides and opposite angles are equal (i.e. BC = AD and ∠A = ∠C).

Consecutive angles are supplementary that is ∠A + ∠B = 180°.

The diagonals of a parallelogram bisect each other also separating the parallelogram into two congruent triangles.

In parallelogram HIJK, ∠H and ∠J are opposite angles. hence:

∠H = ∠J (opposite angles of a parallelogram are equal)

∠J = 45°

8 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

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3 years ago

Find the area of the right triangle.

disa [49]

54 is th answer because ther is 6 units on one side and 9 units in the other and when you times you get this I think

8 0

3 years ago

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At the beginning of each of her four years in college, Miranda took out a new Stafford loan. Each loan had a principal of $5,500

kaheart [24]

Answer:

D. $31,337.27

Step-by-step explanation:

We have that the initial amount of the loan is $5500.

Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.

The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.

Also, it is given that the loan was compounded annually.

We have the formula as,

$P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}$

i.e. $PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}$

Substituting the values, we get,

i.e. $PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}$

i.e. $22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}$

i.e. $22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}$

i.e. $22000=\frac{P\times [1-0.4735]}{0.00625}$

i.e. $22000=\frac{P\times 0.5265}{0.00625}$

i.e. $P=\frac{22000\times 0.00625}{0.5265}$

i.e. $P=\frac{137.5}{0.5265}$

i.e. $P=261.16$

Thus, the total lifetime cost to pay of the loans compounded annually = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

7 0

3 years ago

Read 2 more answers