KJ and KL are the legs of the right triangle below.<br>True or False

Eazy points just answer the question and no I am not a stalker I have a graph to do and theses are the poll question I need 10 p

Setler [38]

6'3

4 sibs

7 times

12 5 20 and 7

8 hours

no

purple

:3 good luck!

7 0

3 years ago

The point nearest to the origin on a line is at (4, -4). Find the standard form of the equation of the line.

Scrat [10]

Just did a specific one of these; let's do the general case.

The point nearest the origin is (a,b).

The line from the origin through the point is

$bx - ay = 0$

The line we seek is perpendicular to this one. We swap the coefficients on x and y, negating one, to get the perpendicular family of lines. We set the constant by plugging in the point (a,b):

$ax + by = a^2 + b^2$

$ax + by -( a^2 + b^2) = 0$

That's standard form; let's plug in the numbers:

$4 x - 4 y - 32 = 0$

$x - y - 8 = 0$

5 0

3 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

Read more about the center of mass.

brainly.com/question/27549055

#SPJ1

8 0

2 years ago

Determine the combined surface area of a cube with an edge that is 3.5 cm long and a cube with an edge that is 2 cm long.

Elis [28]

Answer:

97.5

Step-by-step explanation:

3.5cm cube surface area:

3.5*3.5*6 (the 6 faces of the cube) =

73.5

2cm cube surface area:

2*2*6 =

24

combined surface area:

73.5+24=

97.5

5 0

3 years ago

What is the product?<br><br> (5.63 x 10^5) x 5,600,000

victus00 [196]

Answer: $3.1528 \times 10^{12}.$

Step-by-step explanation: Given expression $(5.63 \times 10^5) \times 5,600,000$ .

In order to find the product of given expression, we need to convert 5,600,000 into scientific notation, because other number is also in scientific notation.

5,600,000 = 5.6 × 10^6.

Therefore, $(5.63 \times 10^5) \times 5,600,000 =(5.63 \times 10^5) \times (5.6 \times 10^6)$ .

Now, multiplying 5.63 × 5.6 first, we get 31.528.

And 10^5 × 10^6 = 10^11.

Therefore,

$(5.63 \times 10^5) \times (5.6 \times 10^6) = 31.528 \times 10^11$ .

6 0

3 years ago

KJ and KL are the legs of the right triangle below.True or False