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3 years ago

Help? SOH, CAH, or TOA

Mathematics

1 answer:

tatyana61 [14]3 years ago

7 0

First you'd find angle C as it is opposite the side AB that you want to find
But to find angle c you first have to find angle A using the soh
Then you'd do 180 - (90 + angle A) =angle C as angles in a triangle add up to 180
Then you'd use the soh using angle A and 12 to find side AB

Working out
Sin(x)=12/7
Sin-1(7/12)
=35.685

180- (90+35.685)
=54.315

Sin(54.315)=o/12
Sin(54.315)*12
=9.746794345

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Which graph shows the solution to the following system y=1\4x-2 y=7\4x+4

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3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

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$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

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$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

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3 years ago

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elixir [45]

Answer:

r = 3

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x^2 + 6x + y^2 - 8y = - 16

Take half the linear term and square it.

x^2 + 6x + (3)^2 + y^2 - 8y + (-4)^2 = - 16

Add the squared amounts to the right.

x^2 + 6x + 9 + y^2 - 8y + 16 = - 16 + 9 + 16

Combine on the right.

x^2 + 6x + 9 + y^2 - 8y + 16 = 9

Represent the 2 quadratics as perfect squares.

(x + 3)^2 + y - 4)^2 = 9

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3 years ago

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victus00 [196]

Answer: A. x ≤ -3

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4 years ago

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Consider the quadratic equation 3x^2-6=2x. (a) What is the value of the discriminant? (b) What does the discriminant of the quad

aleksley [76]

Answer:

(a) discriminant is positive, 76

(b) the above tells you there are two real roots

Step-by-step explanation:

the discriminant is 'b² - 4ac'

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(-2)² - 4(3)(-6) = 4 + 72

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3 years ago