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3 years ago

12

Can you please help me with number one and number 2 thank you

2 answers:

maksim [4K]3 years ago

6 0

2+19-7=14 i believe is for #1

Vlad [161]3 years ago

3 0

1 question: the end placement would be 9 feet
-1 ft
START-2 ft
-3 ft
-4 ft
-5 ft
-6 ft
-7 ft
-8 ft
END-9 ft
-10 ft
-11 ft
-12 ft
-13 ft
-14 ft
-15 ft
-16 ft
-17 ft
-18 ft
where it goes first-19 ft

2nd question: 45-8= 37
37+53= 90
90-6= 84

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Stolb23 [73]

Answer:

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Step-by-step explanation:

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3 years ago

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A 3x4x5 rectangular cuboid (e.g. a brick) is painted blue and cut into 1x1 cubes. what is the expected value for the painted sid

Anton [14]

Answer:

8 cubes are painted on 3 sides

24 cubes are painted on 2 sides

22 cubes are painted on 1 sides

6 cubes are painted on 0 side

E = 3 * 8/60 + 2 * 24/60 + 1 * 22/60 + 0 * 6/60 = 47/30 = 1.567

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3 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

m_a_m_a [10]

The value of the card was $13 when he originally purchased it.

7 0

3 years ago

NEED HELP NOWWW Which of the following is a monomial?

Dafna11 [192]

Answer: C

Step-by-step explanation:

A monomial is a expression where in it is x to the power of something, and x cannot be a denominator

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3 years ago