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3 years ago

9

find three consecutive even integers such that the sum of the least integer and the middle integer is 22 more than the greatest

integer

1 answer:

mrs_skeptik [129]3 years ago

7 0

Your integers are x, x+2, and x+4
x+x+2=22+x+4
2x+2=x+26
x=24
the integers are 24,26,28

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What additional information do you need to prove that ∆AMD ≅ ∆CNB by the HL Theorem?

Maurinko [17]

(H)ypotenuse (L)eg, right-triangle theorem.

based on the provided graph, we know the hypotenuses are equal, thus the tickmarks, so H is true, now the L part, we only need either of the pair of legs to be equal DM = NB or AM = CN, if either one is true, we're golden.

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3 years ago

Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

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3 years ago

A class of 12 girls and 14 boys is going to choose 3 students to represent them in student government. Allie suggested that the

Vikki [24]

<h2>Answer:</h2>

The method suggested by Bobby is better than Allie.

<h2>Step-by-step explanation:</h2>

A fair sample is a sample where each sample has an equal opportunity to get selected for any survey.

Here, in the given question, Allie suggested that the first 3 students to arrive in class the next day should be the representatives. This sample is not random or fair.

Whereas, Bobby suggested that each student roll 2 numbers cubes and the 3 students with the highest sum should be the representatives. This method is a fair method.

The method suggested by Bobby is better than Allie as every student will get an equal chance to perform and also has an equal chance to win and get a high sum.

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3 years ago

Read 2 more answers

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tigry1 [53]

The vector <em>v</em> (-6, 6, -5) points in the direction of itself, so we start there.

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The line through point <em>P</em> and pointing in the same direction as <em>v</em> is parallel to the other line that passes through the origin. Then the line we want can be obtained by translating the line through the origin by a vector <em>p</em> that points to (-4, 5, 2), so the vector equation for this line is

<em>r </em>(<em>t </em>) = <em>p</em> + <em>t v</em>

<em>r </em>(<em>t </em>) = (-4, 5, 2) + <em>t</em> (-6, 6, -5)

<em>r </em>(<em>t </em>) = (-4 - 6<em>t</em>, 5 + 6<em>t</em>, 2 - 5<em>t</em> )

To get the parametric equations, simply take out the components:

<em>x</em> (<em>t</em> ) = -4 - 6<em>t</em>

<em>y </em>(<em>t</em> ) = 5 + 6<em>t</em>

<em>z</em> (<em>t</em> ) = 2 - 5<em>t</em>

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3 years ago

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Answer:

4 and 4 inches left

Step-by-step explanation:

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3 years ago