Multiply or find the special product. (Simplify your answer completely.) [(x − 8y) + z][(x − 8y) − z]

1 answer:

3 0

<span>[(x − 8y) + z] * [(x − 8y) − z]

= </span>(x − 8y)² - z²
[ Following the rule: a² - b² = (a + b)(a - b) where 'a' here is (x − 8y) and 'b' is z ]

= x² -16xy +64y² - z²

= x² + 64y² - z² - 16xy

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Complemetary angles are two angles whose measures add up to 90 degrees. The ratio of the measures of two complementary angles is

pochemuha

Let A and B be the two complementary angles.

A = smaller angle = 2x

B = larger angle = 13x

x = some unknown number

Note how the ratio A:B turns into 2x:13x which simplifies to 2:13

A+B = 90 ... because the angles are complementary

2x+13x = 90 ... substitution

15x = 90

x = 90/15

x = 6

A = 2*x = 2*6 = 12 degrees

B = 13*x = 13*6 = 78 degrees

The two angles are 12 degrees and 78 degrees.

Check:

A/B = 12/78 = (2*6)/(13*6) = 2/13, so A:B = 2:13

A+B = 12+78 = 90

3 0

3 years ago

Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis

Tasya [4]

Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5

A.
$\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx$

B.
sepearte the integrals
$\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}$

next one
$\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}$

the last one you can do yourself, it is $\frac{50}{3}$
the sum is $\frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}$

so the area under the curve is $\frac{233}{3}$

6 0

3 years ago

The surface areas of two similar solids are 340 yd² and 1,158 yd². The volume of the larger solid is 1,712 yd³. What is the volu

ratelena [41]

The ratio of surface area is equal to the ratio of the square of the corresponding dimensions. And ratio of volumes of two solids is equal to the cube of the ratio of the corresponding dimensions .

We start with the relation between ratio of surface area and ratio of corresponding sides. That is

$\frac{340}{1158}=(\frac{x}{y} )^2$