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2 years ago

Passes through (1,4) and is parallel to the line y=2x-3

Mathematics

1 answer:

Novay_Z [31]2 years ago

3 0

Answer:

y = 2x + 2

Step-by-step explanation:

Slope = 2

y-intercept: 4 - (2)(1) = 2

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The isosceles trapezoid is part of an isosceles triangle with a 32° vertex angle. What is the measure of an acute base angle of

nevsk [136]

Answer:

acute base angle = 74° , obtuse base angle = 106°

Step-by-step explanation:

Since the large triangle is isosceles with vertex angle (top angle) is 32, then the bottom 2 angles would be same (let it be x):

We know angles of triangle add up to 180, so we have:

x + x + 32 = 180

2x = 180 - 32

2x = 148

x = 148/2

x = 74

This 74 degree angle is the base acute angle of the isosceles trapezoid (lower portion). We also know opposite angles of isosceles trapezoid are supplementary (add up to 180), thus

obtuse angle + 74 = 180

obtuse angle (base) = 180 - 74 = 106

Thus, acute base angle = 74° , obtuse base angle = 106°

6 0

3 years ago

To find the quotient Two-fifths divided by one-fourth,

lutik1710 [3]

Answer:

this answer has to be A because the libra of the pole goes west

Step-by-step explanation:

this answer has to be A because the libra of the pole goes west

6 0

3 years ago

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Juli2301 [7.4K]

You’re correct answer would be : 14 lmk if this is correct if not I’m so SORRY

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3 years ago

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Vika [28.1K]

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Step-by-step explanation:

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3 years ago

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Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago