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3 years ago

What is an equation of the line that passes through the points (-1, 6) and (-1, -5)?

Mathematics

1 answer:

kobusy [5.1K]3 years ago

4 0

Answer:

x = -1

Step-by-step explanation:

(-1 , 6) ; (-1 , -5)

There is no change in x-coordinate values.

So, the is parallel to y-axis

x = -1

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Write a word problem that can be solved by ordering three decimals to thousandths. Include a solution

nikklg [1K]

Then make up a word problem that you can use decimals in.

6 0

3 years ago

Bulan rows on a crew team. Her team rows their boat at a split (rate) of 2 min/500 m.

notka56 [123]

Answer:

250

Step-by-step explanation:

If Bulan's team rows their boat at a rate of

minutes per

500

meters, they row at a rate of

minute per

250

meters. We know this because

minute is

minutes, and in this time, they will have to have rowed

the distance they would row in

minutes (

500

m).

500

250

So, we now have the rate in min/m.

If, every

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250

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Bulan's team's rowing rate in m/min is

250

min.

8 0

3 years ago

Read 2 more answers

-8u = -7u+ 5<br> U=<br> Help

Thepotemich [5.8K]

Answer:

U = -5

Step-by-step explanation:

-8u = -7u + 5

-u = 5

u = -5

3 0

2 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

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

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago

Evaluate: 2|x-4|+6 for x=2

Lina20 [59]

<h3>Answer:</h3>

$10$

<h3>Step-by-step explanation:</h3>

$2|x -4| +6 \text{ for } x = 2 \\ 2|2 -4| +6 \\ 2|-2| +6 \\ 2(2) +6 \\ 4 +6 \\ 10$

6 0

3 years ago