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2 years ago

9

Select all the expressions that are equivalent to 13 - x *

1 answer:

alina1380 [7]2 years ago

7 0

You gave no options but the simplest way to solve this is to set it equal to all of the answers to see if it is correct.

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What is seven and one fifth minus four and five eighths?<br> 01 7/8<br> 2 23/40<br> 3 3/8<br> 4 1/4

dalvyx [7]

Answer:

(7+1÷5)−(4+5÷8) = 2 23/40 B

Step-by-step explanation:

7 1/5 - 4 5/8

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2 years ago

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Complete the following statement. Round to the nearest hundredth .

RSB [31]

f(x) = 3/(x + 2) - √x - 3

f(7) = 3/(7 + 2) - √7 - 3

f(7) = 3/9 - √4

f(7) = 0.333 - 2

f(7) = -1.67

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2 years ago

Based on the table, which best predicts the end<br> behavior of the graph of f(x)?

Iteru [2.4K]

Answer:

approaching negative infinity

Step-by-step explanation:

Since as x increases, the values of f(x) are approaching infinity, the function approaches negative infinity as the end behavior.

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2 years ago

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Nicky has 5 cookies and bill has 4 cookies how much do they have all together

Vsevolod [243]

9 cookies

5 plus 4 is 9

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2 years ago

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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

2 years ago