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vampirchik [111]

2 years ago

7

A segment has endpoints A (-1, 1) and B (8, 4) . If the segment is divided into four equal parts, the coordinates of the point c

losest to point A are _____. (7/2, 5/2) (5/4, 7/4) (23/4, 13/4)

2 answers:

marin [14]2 years ago

7 0

Answer:

ANSWER IS ( 5/4 , 7/4 ).

Step-by-step explanation:

riadik2000 [5.3K]2 years ago

6 0

Line is divided into 4 equal parts.

we have to find a point which is closest to point A.

So that means required point P(x,y) is at 1 unit away from A(-1,1) and 3 unit away from B(8,4)

Now we just need to use section formula to get the coordinate of required point using m1=1 and m2=3

$\left ( \frac{m_1x_2+m_2x_1}{m_1+m_2} , \frac{m_1y_2+m_2y_1}{m_1+m_2} \right )$

$= \left ( \frac{1*8+3*(-1)}{1+3} , \frac{1*4+3*1}{1+3} \right )$

$= \left ( \frac{8-3}{4} , \frac{4+3}{4} \right )$

$= \left ( \frac{5}{4} , \frac{7}{4} \right )$

So the final answer is $\left ( \frac{5}{4} , \frac{7}{4} \right )$ .

You might be interested in

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

-6(1 + 2v) – 7(v + 1) = -13

docker41 [41]

The answer is V=0.3157=0.32

6 0

2 years ago

zheka24 [161]

Step-by-step explanation:

${x}^{65559$

8 0

2 years ago

Help I’m being timed!!

jok3333 [9.3K]

Trueeeeeeeeeeeeeeeeeeee

6 0

3 years ago

Read 2 more answers

The graph of a quadratic function has a vertex at the point (8,6). It passes through the point (-4,-5). Which of the following e

Mandarinka [93]

Answer:

y= -11/144(x-8)^2+6

Step-by-step explanation:

This equation can be represented in vertex form, which is:

y=a(x-h)^2+k

If we plug in 8 as h and 6 as k we get the following equation:

y=a(x-8)^2+6

Now we have to plug in x and y. We can use the other point (-4,-5) and plug it into the equation and get:

-5=a(-4-8)^2+6

Once we solve this we get a= -11/144

Now we have to plug in -11/144 into the original equation to get

y = -11/144 (x-8)^2 + 6

6 0

2 years ago