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3 years ago

Please help! Remember what we know about vertical angles and solve for x. Answer:

Mathematics

1 answer:

aleksklad [387]3 years ago

4 0

Please help!
Remember what we know about vertical angles and solve for x.
Answer:

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PLEASE HELPPPPPP!!!!!

Anarel [89]

The Answer Is (A)

y equals a positive 2x

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3 years ago

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PLEASE HELP ASAP, I WILL GIVE BRAINLESSLY ANSWER <br> SHOW WORK PLEASE

zhannawk [14.2K]

Answer:

C. $N(t)=150\cdot 3^t$

Step-by-step explanation:

You are given the exponential function $n(t)=ab^t.$

From the table, $N(t)=150$ at $t=0,$ thus

$N(0)=a\cdot b^0\\ \\150=a\cdot 1\ [\text{ because }b^0=1]$

Also $N(t)=450$ at $t=1,$ thus

$N(1)=a\cdot b^1=a\cdot b.$

Since $a=150,$ substitute it into the second equation

$450=150\cdot b\\ \\b=\dfrac{450}{150}\\ \\b=3$

and the expression for the exponential function is

$N(t)=150\cdot 3^t$

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3 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

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3 years ago

Which correctly renames

Goshia [24]

Answer:

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Step-by-step explanation:

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2 years ago

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VERY EASY, WILL GIVE 50 POINTS FOR CORRECT ANSWER ASAP AND WILL GIVE BRAINLIEST.

timurjin [86]

Answer:

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3 years ago

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