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3 years ago

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Answer Please! Will give Brainlist! 'Easy'

1 answer:

Nataly [62]3 years ago

6 0

Answer:

i think its 0.4x+7<1 I don't really know

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X/2-1/5=x/3+1/4<br>solve this equation

Alekssandra [29.7K]

Answer:

$x=\frac{27}{10}$

Step-by-step explanation:

$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

$\frac{x}{2}-\frac{1}{5}-\frac{x}{3}=\frac{x}{3}+\frac{1}{4}-\frac{x}{3}$

$-\frac{1}{5}+\frac{x}{6}=\frac{1}{4}$

$-\frac{1}{5}+\frac{x}{6}+\frac{1}{5}=\frac{1}{4}+\frac{1}{5}$

$\frac{x}{6}=\frac{9}{20}$

$\frac{6x}{6}=\frac{9\times \:6}{20}$

$x=\frac{27}{10}$

3 0

3 years ago

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If the farm has 30 chickens and cows, and there are 82 chicken and cow legs all together, then how many chickens and how many co

natima [27]

There would be 11 cows and 19 chickens

3 0

3 years ago

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Please solve the following sum or difference identity.

xxTIMURxx [149]

Answer:

$sin(A - B) = \frac{4}{5}$

Step-by-step explanation:

Given:

$sin(A) = \frac{24}{25}$

$sin(B) = -\frac{4}{5}$

Need:

$sin(A - B)$

First, let's look at the identities:

sum: $sin(A + B) = sinAcosB + cosAsinB$

difference: $sin(A - B) = sinAcosB - cosAsinB$

The question asks to find sin(A - B); therefore, we need to use the difference identity.

Based on the given information (value and quadrant), we can draw reference triangles to find the simplified values of A and B.

sin(A) = $\frac{24}{25}$

cos(A) = $\frac{7}{25}$

sin(B) = $-\frac{4}{5}$

cos(B) = $\frac{3}{5}$

Plug these values into the difference identity formula.

$sin(A - B) = sinAcosB - cosAsinB$

$sin(A - B) = (\frac{24}{25})(\frac{3}{5}) - (-\frac{4}{5})(\frac{7}{25})$

Multiply.

$sin(A - B) = (\frac{72}{125}) + (\frac{28}{125})$

Add.

$sin(A - B) = \frac{4}{5}$

This is your answer.

Hope this helps!

6 0

3 years ago

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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):

<u>Points (0,0) and (2,1):</u>

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

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

<u>Points (2,1) and (0,3):</u>

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

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

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:

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

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

$M_{x}$ and $M_{y}$ are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

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

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

$M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2} {y.x+y^{2}} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24} )}\, dx }$

$M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)$

$M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)$

$M_{x} = 18$

Now to find the x-coordinate:

x = $\frac{M_{y}}{m}$

x = $\frac{63}{4}$

x = 15.75

For moment about the y-axis:

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

$M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {x^{2}+yx} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2} } \,dx }$

$M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})$

$M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)$

$M{y} =$ 63

To find y-coordinate:

y = $\frac{M_{x}}{m}$

y = $\frac{18}{4}$

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0

3 years ago

What percent of 200 miles is 150 miles

Bumek [7]

150 / 200 = 0.75

0.75 = 75%

150 of anything is 75% of 200 of the same thing.

3 0

3 years ago

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