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3 years ago

Jess is painting his car. He's created a purple color that uses red paint and blue paint in a ratio of 6:5. Jess uses 15

Mathematics

2 answers:

lilavasa [31]3 years ago

4 0

B
I’m sure It’s current answer

Inessa [10]3 years ago

4 0

Answer:

B.) 33

Step-by-step explanation:

15/5=3

6x3=18

18+15=33

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What is the exact volume of the cylinder enter your answers in the term of pie in the Box 5cm 15cm

olga_2 [115]

Answer:

Valume of cylinder : πr² × h = π × 5² × 15 = 375π ≈ 1177.5 cm³

8 0

3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x+2}$ if it is defined for $x = 0$ .

Answer:

$y = \sqrt {x-2}$

Option b

6 0

3 years ago

I need help finding the first five terms of the following explicit formula and the sequence

Katarina [22]

Answer:

Step-by-step explanation:

Explicit formula of the sequence,

$A_n=59+11n$

$A_1=59+11(1)$

$A_1=70$

$A_2=59+11(2)$

$A_2=81$

$A_3=59+11(3)$

$A_3=92$

$A_4=59+11(4)$

$A_4=103$

$A_5=59+11(5)$

$A_5=114$

8 0

3 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

Points (0,0) and (2,1):

y = $\frac{1-0}{2-0}(x-0)$

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

Points (2,1) and (0,3):

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

Mass of the lamina that occupies region D is 4.

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$