Ask question

Ask question

All categories

pickupchik [31]

2 years ago

15

Multiply 11 x 34.12 I’m so confused?

1 answer:

garik1379 [7]2 years ago

5 0

Answer:

$34.12 * 11 = 375.32$

Step-by-step explanation:

Given

11 * 34.12

Required

Multiply

To start with, we have to ignore the decimal points;

So, we have

3412

* 11

---------------

3 4 1 2

3 4 1 2

------------------------

3 7 5 3 2

The next step is as follows;

34.12 is in 2 decimal places

11 is has no decimal (0 decimal place)

Add the decimal places together

$Decimal\ Place = 2 + 0 = 2$

This implies that the result will be in 2 decimal placed;

Hence:

$34.12 * 11 = 375.32$

You might be interested in

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

Nuance said she drew a square pyramid and that all of the faces are triangles.Is this possible?

sergeinik [125]

No because a square pyramid must hav a square for its base and a square is not a triangle

answer is not possible

3 0

2 years ago

Read 2 more answers

Multiply 2(x+3)/x(x-1)*4x(x+2)/10(x+3)

KIM [24]

Answer:

The expression simplifies to $\frac{4(x+2)}{5(x-1)}$ .

Step-by-step explanation:

The expression

$\frac{2(x+3)}{x(x-1)} *\frac{4x(x+2)}{10(x+3)}$

can be rearranged and written as

$\frac{8x(x+3)(x+2)}{10x(x-1)(x+3)}.$

In this form the $(x+3)$ terms in the numerator and in the denominator cancel to give

$\frac{8x(x+2)}{10x(x-1)}.$

The $x's$ are present both in the numerator and in the denominator, so they also cancel, and the fraction $\frac{8}{10}$ simplifies to $\frac{4}{5}$ , so finally our expression becomes:

$\boxed{\frac{4(x+2)}{5(x-1)}}$

Which is our answer:)

6 0

2 years ago

Find the equation of the line passing through the points (1,5) and (3,9)

Arada [10]

Answer:

$\boxed{\sf{2}}$

Step-by-step explanation:

Use the slope formula.

$\underline{\text{SLOPE:}}$

$\Longrightarrow: \sf{\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{Rise}{Run} }$

$\sf{y_2=9}\\\\\\\sf{y_1=5}\\\\\\\sf{x_2=3}\\\\\\\sf{x_1=1}$

$\sf{\dfrac{9-5}{3-1} }$

Solve.

$\sf{\dfrac{9-5}{3-1}=\dfrac{4}{2}=\boxed{\sf{2}}$

<u>Therefore, the slope is 2, which is our answer.</u>

I hope this helps. Let me know if you have any questions.

8 0

1 year ago

What the constant proportionality of y=26x

UNO [17]

Answer:

Here is a link to help you. This is a very great video that will help you understand proportionalities .

Step-by-step explanation:

https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-ratio-proportion/7th-constant-of-proportionality/v/constant-of-proportionality-from-equation

5 0

2 years ago