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3 years ago

Find the equation of the line. Use exact numbers.

Mathematics

1 answer:

postnew [5]3 years ago

3 0

Answer:

y=1/2x-3

is the slope intercept form

Step-by-step explanation:

Hope this helps :) Have a great day!!!

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The area of a triangle is one half the base and times the height. If the area of the triangle is 16 square inches and the vase i

sergij07 [2.7K]

Step-by-step explanation:

so, yes, the area of a triangle is

baseline × height / 2

so, we have here

16 = 8 × height / 2 = 4 × height

height = 16/4 = 4 in.

8 0

2 years ago

What is .744 as a fraction

jenyasd209 [6]

744/1000 would be your answer

4 0

3 years ago

Read 2 more answers

Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.

tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: $\frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3}$ square units

(22)

Similar to (21)

Area = $216\sqrt{3}$ square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

$altitude=\frac{\sqrt{3}}{2}*side$

$side = \frac{36}{\sqrt{3}}$

Hence, area of the hexagon will be: $648\sqrt{3}$ square units

(24)

Given is the inradius of an equilateral triangle.

$Inradius = \frac{\sqrt{3}}{6}*side$

Substituting the value of inradius and calculating the length of the side of the equilateral triangle:

Side = 16 units

Area of equilateral triangle = $\frac{\sqrt{3}}{4}*(side)^{2} = \frac{\sqrt{3}}{4}*256 = 64\sqrt{3}$ square units

4 0

3 years ago

I need help ASAP!!!!!!!!!!

Lorico [155]

Answer:

Step-by-step explanation:

7 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago

Find the equation of the line. Use exact numbers. ​

Find the equation of the line. Use exact numbers.