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3 years ago

What must be true for lines a and b to be parallel lines? Check all that apply.

Mathematics

2 answers:

Dima020 [189]3 years ago

5 0

Answer:

Option (1), (2) , (7) and (8) is correct.

Step-by-step explanation:

Given: a║b and d and c be transversal , We have to find the value of x and measure of angle 1 and 2.

Since, a║b and d be a transversal then ,

∠BXZ = ∠AZP = 58° ( corresponding angles )

Thus, m∠2 = 58°

∠BYC = ∠XYZ = (4X-10)° (vertically opposite angles)

Since, a║b and c be a transversal then,

∠XYZ = ∠QZU = (4X-10)° (Corresponding angles)

Thus, m∠1 = (4X-10)°

Also, Since, 'a' is a straight line.

Sum of angles on a straight line is 180°.

⇒ ∠PZA + ∠PZQ + ∠QZU = 180°

⇒ 58° +(3X-1)°+(4X-10)° = 180°

⇒ 58° + 7x - 11 = 180°

⇒ 47° + 7x = 180°

⇒ 7x = 180°- 47°

⇒ 7x = 133°

⇒ x = 19°

Put x = 19 in (3X-1)° and (4X-10)° , we get

(3X-1)° = (3(19)-1)° = 56°

(4X-10)° = (4(19)-10)°= 66

Thus, (3X-1)° ≠ (4X-10)°

Thus, option (1), (2) , (7) and (8) is correct.

a_sh-v [17]3 years ago

5 0

I think I got most of it.

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Evaluate triple integral

kaheart [24]

Answer:

$\\ \frac{1}{8} e^{4a}-\frac{3}{4}e^{2a}+e^{a} -\frac{3}{8} \\\\or\\\\ \frac{e^{4a}-6e^{2a}+8e^{a}-3}{8}$

Step-by-step explanation:

$\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\$

$\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\$

$\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\$

$\\=\frac{1}{8}e^{u_1}\Big| -\frac{3}{4}e^{u_2}\Big| + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x}\Big|_{0}^a -\frac{3}{4}e^{2x}\Big|_{0}^a + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x} -\frac{3}{4}e^{2x} + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{1}{8} +\frac{3}{4} -1\\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{3}{8}\\\\\\$

Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.

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UkoKoshka [18]

Answer:

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The function reaches its maximum height at ...

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The function can be factored to find the zeros.

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The factors are zero for ...

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dexar [7]

Answer:

x = 2root(22)

see image.

Step-by-step explanation:

The triangle shown is one big triangle cut into two more smaller triangles: one medium-sized and one smaller.

ALL THREE TRIANGLES ARE SIMILAR BY AA.

Set the two smaller triangles up so you can see the corresponding sides. x is the short leg in one triangle and it is the long leg in the smallest triangle. Set up a proportion.

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