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2 years ago

Help please!!! I only have 10 minutes!!

Mathematics

2 answers:

kiruha [24]2 years ago

4 0

$- 0.7$

Step-by-step explanation:

$\frac{1}{2} ( - 1.4m + 0.4)$

$= - 0.7m + 0.2$

goblinko [34]2 years ago

3 0

I think 0.7 would be it

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A line has the equation -5x -14y +168 = 0

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Answer:

y = -5/14x + 12

Step-by-step explanation:

add 14y to both sides to isolate variables and divide everything by 14 to get 1y and boom

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2 years ago

What is the value of the expression (2 x 5)^3

Tom [10]

Deal with the brackets first
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Then 10 cubed = 1000

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3 years ago

Triangle ABC had a perimeter 18 cm

Anna007 [38]

Answer:

Use pythagoreum theorem to determine whether it is or isn't since I'm not really sure myself but here's my calculations.

Step-by-step explanation:

We need to find AC.

Picture a triangle called ABC. AB is one side and BC the other while AC is the hypotenuse.

AC=

$ac = \sqrt{ab ^{2} + {bc}^{2} } \\ ac = \sqrt{ {7}^{2} } + {3}^{2} \\ ac = \sqrt{49 + 9} \\ ac = \sqrt{58 } \\ ac = 7.6$

Round up the 7.6 to a whole number and you get AC=8. To get perimeter of a triangle simply add the sides together to get 18.

Not really sure about my answer but hope this helped.

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2 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$