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3 years ago

Find x if m<SGH = 160o, m<FGH = 46x - 4, and m<FGS = 6x - 4 Please help me with this math problem

Mathematics

1 answer:

Ugo [173]3 years ago

4 0

Once you draw out the angles (see my attached picture). you should notice that m<FGH + m<FGS = 160. We can then set this up as an equation.

m<FGH + m<FGS = 160
(46x-4)+(6x-4)=160
52x-8=160
52x=168
x is approximately 3.23.

:)

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Norma-Jean [14]

Answer:

x = 28 cm

Step-by-step explanation:

Given:

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Required:

The value of x (diameter of the semicircle/length of the rectangle)

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Length of rectangle (L) = 2r = x

Radius of semicircle (r) = ½x

Width of rectangle (W) = radius of semicircle = ½x

Use 3.14 as π

Area of the link shaded regions = area of rectangle - area of semicircle

Thus:

Area of the link shaded regions = (L*W) - (½*πr²)

Plug in the values

84 = (x*½x) - (½*3.14*(½x)²)

84 = x²/2 - (1.57*x²/4)

84 = x²(½ - 1.57/4)

84 = x²(0.5 - 0.3925)

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Divide both sides by 0.1075

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2 years ago

Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

MAVERICK [17]

Answer:

a. $\frac {2} {s-1}$ converges to s> 1.

b. $\frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $- \frac {2}{s + 3}$ converges to s> - 3.

d. $\frac {s}{s^2 + 25}$ converges to s> 0.

e. $\frac {10} {s^2 + 1}$ converges even s> 0.

f. $\frac {12}{s^2 + 4}$ converges to s> 0.

g. $-\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $\frac {1} {s ^ 2 + 4}$ converges to s> 0.

Step-by-step explanation:

a. $L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1}$ converges to s> 1.

b. $L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3}$ converges to s> - 3.

d. $L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25}$ converges to s> 0.

e. $L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1}$ converges even s> 0.

f. $L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4}$ converges to s> 0.

g. $L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4}$ converges to s> 0.

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3 years ago

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2 years ago

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lana66690 [7]

Answers

1) 1

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Explanation

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Slope = (change in y)/(change in x)

= (0 - -2)/(-4 - -6)

= 2/2

= 1

Slope = Δy/Δx

= (8 - -3)/(2 - 1)

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Y ∞ X

Put and equal sign then introduce the constant of proportionality, K.

Y = KX

K = Y/X

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∴ Y = (1/3)X

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Dividing both sides by 32,

Y = 96/32 X

Y = 3X

3 been the constant, Y varies directly as X.

So, the equation 32y = 96x is a direct variation.

Constant of variation = 3

5 0

3 years ago

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sesenic [268]

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2 years ago