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3 years ago

Solve for x. −8 = 2/3x A) −6 B) −9 C) −12 D) −24

Mathematics

1 answer:

Reil [10]3 years ago

7 0

Answer:

C. -12

Step-by-step explanation:

divide each side by 2/3

2/3x ÷ 2/3 = x

-8 ÷ 2/3 = -12

x = -12

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Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in t

Tresset [83]

Answer: The volume of largest rectangular box is 4.5 units.

Step-by-step explanation:

Since we have given that

Volume = $xyz$

with subject to $x+2y+3z=9$

So, let $z=\dfrac{9-x-2y}{3}$

So, Volume becomes,

$V=xyz\\\\V=xy(\dfrac{9-x-2y}{3})\\\\V=\dfrac{9xy-x^2y-2xy^2}{3}$

Partially derivative wrt x and y we get that

$9-2x-2y=0\implies 2x+2y=9\\\\and\\\\9-x-4y=0\implies x+4y=9$

By solving these two equations, we get that

$x=3,y=\dfrac{3}{2}$

So, $z=\dfrac{9-x-2y}{3}=\dfrac{9-3-3}{3}=\dfrac{3}{3}=1$

So, Volume of largest rectangular box would be

$xyz=3\times \dfrac{3}{2}\times 1=\dfrac{9}{2}=4.5$

Hence, the volume of largest rectangular box is 4.5 units.

4 0

3 years ago

What is 160% of 5 rounded to the nearest hundredth

faltersainse [42]

Answer:

Step-by-step explanation:

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3 years ago

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leva [86]

Answer:

Step-by-step explanation:

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3 years ago

Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum

lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

$\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\$

$\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\ (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\$

$=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3} \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\$

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2 years ago

Find the coordinates of the center and the measure of the radius for a circle whose equation is (x-3)^2 + (y-7)^2=2

igomit [66]

Answer:

fourth option

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x - 3)² + (y - 7)² = 2 ← is in standard form

with centre = (h, k) = (3, 7) and r² = 2 ⇒ r = $\sqrt{2}$

6 0

3 years ago