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3 years ago

15

Let r be the region bounded by y = x 2 and y = 9. find a and b so that the horizontal lines y = a and y = b split the region r i

nto 3 pieces of equal area.

1 answer:

Arturiano [62]3 years ago

8 0

y = x² has a lower bound at the vertex (0, 0); <em> y = 0</em>

y = 9 has an upper bound at y = 9

x² = 9 ⇒ x = +/- 3 the region is bounded on the left at x = -3 and on the right at x = 3

Solve the integral from -3 to 3 to find the area under the curve, then multiply by 1/3 to calculate 1/3 of the area.

$\int\limits^3_3 (9 - {x^{2}}) \, dx$ = (9x - x³/3) | from -3 to 3

= (9(3) - 3³/3) - (9(-3) - (-3)³/3)

= (27 - 9) - (- 27 - (-9))

= 16 - (-16)

= 32

*** 32/3 is 1/3 of the area ****

Now that we know what 1/3 of the area is, we can find the height <em>at y = a</em> which will satisfy 1/3 of the area. We do this by solving the integral from "0 to a" and setting it equal to 1/3 of the area (32/3).

$\int\limits^a_0 (9 - {x^{2}}) \, dx$ - a(9 - a²)

= (9x - x³/3) | from 0 to a - (9a - a³)

= (9a - a³/3) - (9(0) - (0)³/3) - 9a + a³

= 9a - a³/3 - 9a + a³

= a³ - a³/3

= 2a³/3

32/3 = 2a³/3

32 = 2a³

16 = a³

∛16 = a

2∛2 = a (This is the distance from the upper bound (y = 9)

Answer: y = 9 - 2∛2

Follow the same steps using 2/3 of the area (64/3) to find "b".

64 = 2b³

32 = b³

∛32 = b 3.17

2∛4 = b (This is the distance from the upper bound (y = 9)

Answer: y = 9 - 2∛4

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3 years ago

Help based offf order of operations please help

JulijaS [17]

Answer:

The answer for :

$h. \: \: \: \frac{5}{6}$

$i. \: \: \: \frac{35}{32}$

$k. \: \: \: \frac{-29}{20}$

$l. \: \: \: \frac{13}{15}$

Step-by-step explanation:

Question h:

$\frac{2}{3} + ( \frac{1}{3} \times \frac{1}{2} )$

$= \frac{2}{3} + \frac{1}{6}$

$= \frac{2 \times 2}{3 \times 2} + \frac{1}{6}$

$= \frac{4}{6} + \frac{1}{6}$

$= \frac{5}{6}$

Question i:

$\frac{7}{8} + \frac{1}{4} \times ( \frac{3}{2} - \frac{5}{8} )$

$= \frac{7}{8} + \frac{1}{4} \times ( \frac{3 \times 4}{2 \times 4} - \frac{5}{8} )$

$= \frac{7}{8} + \frac{1}{4} \times ( \frac{12}{8} - \frac{5}{8} )$

$= \frac{7}{8} + (\frac{1}{4} \times \frac{7}{8} )$

$= \frac{7}{8} + \frac{7}{32}$

$= \frac{7 \times 4}{8 \times 4} + \frac{7}{32}$

$= \frac{28}{32} + \frac{7}{32}$

$= \frac{35}{32}$

Question k:

$\frac{3}{4} - ( \frac{12}{7} \div \frac{12}{21} ) + \frac{4}{5}$

$= \frac{3}{4} - ( \frac{12}{7} \times \frac{21}{12} ) + \frac{4}{5}$

$= \frac{3}{4} - \frac{3}{1} + \frac{4}{5}$

$= \frac{3 \times 5}{4 \times 5} - \frac{3 \times 20}{1 \times 20} + \frac{4 \times 4}{5 \times 4}$

$= \frac{15}{20} - \frac{60}{20} + \frac{16}{20}$

$= - \frac{29}{20}$

Question l:

$\frac{5}{2} \times ( \frac{2}{3} - \frac{1}{5} ) - ( \frac{2}{5} \div \frac{4}{3} )$

$= \frac{5}{2} \times ( \frac{2 \times 5}{3 \times 5} - \frac{1 \times 3}{5 \times 3} ) - ( \frac{2}{5} \times \frac{3}{4} )$

$= \frac{5}{2} \times ( \frac{10}{15} - \frac{3}{15} ) - \frac{3}{10}$

$= (\frac{5}{2} \times \frac{7}{15}) - \frac{3}{10}$

$= \frac{7}{6} - \frac{3}{10}$

$= \frac{7 \times 5}{6 \times 5} - \frac{3 \times 3}{10 \times 3}$

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8 0

3 years ago

So here's the question:

Kobotan [32]

Answer:

The angle θ, that the device needs to scan is 16.795°

Step-by-step explanation:

Here we have

Height of room = 11 ft

Width of room = 15 ft

Length of room = 18 ft

Position of camera = 6 inches below the ceiling = 11 ft - 6 in = 126 in = 10.5 ft

Distance from camera to edge of long side of the room is given by the following relation;

Long edge of angle = √((18 ft)²+(10.5 ft)²) = 20.839 ft

Shorter edge of angle = √((15 ft)²+(10.5 ft)²) = 18.31 ft

Opposite side of required angle = √((18 ft)²+(15 ft)²) = 23.431 ft

Therefore, by cosine rule, we have

a² = b² + c² - 2·b·c·cos A

We therefore put our a as the opposite side of the required angle, A so we can easily solve for it

our b and c are then the other two sides

23.431² = 20.839² + 18.31² - 2×20.839×18.31×cosA

∴ cos(A) = (23.431² - (20.839² + 18.31²))÷(2×20.839×18.31)

cos(A) = 220.5/763.12418 = 0.29

A = cos⁻¹0.29 = 16.795°

The angle θ, that the device needs to scan = 16.795°.

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3 years ago

what is 432,010.1 in word form and nine hundred forty-five and thirty-five hundredths in standard form and seventy-three thousan

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1. Four hundred thousand thirty two and ten point one

2. Standard form is adding it out ( 945.35 ) = 900 + 40 + 5 + .30 + .05

3. Expanded form ( 73,890 )

| ten thousands | thousands | hundred | tens | ones | decimal | tenths | hundredths |

7 3 8 9 0 . 0 1

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3 years ago

Add one term to the polynomial expression 14x^19 - 9x^15 + 11x^4 + 5x^2 + 3 to make it into a 22nd degree polynomial

GuDViN [60]

Given:

The polynomial is

$14x^{19}-9x^{15}+11x^4+5x^2+3$

To find:

One term which is used to add in given polynomial to make it into a 22nd degree polynomial.

Solution:

Degree of a polynomial is the highest power of the variable.

Let,

$P(x)=14x^{19}-9x^{15}+11x^4+5x^2+3$

Here, the highest power of x is 19, so degree of polynomial is 19.

To make it into a 22nd degree polynomial, we need to need a term having 22 as power of x.

We can add $kx^{22}$ , where k is constant.

So add $x^{22}$ in the given polynomial.

$P(x)=x^22+14x^{19}-9x^{15}+11x^4+5x^2+3$

Now, the degree of polynomial is 22.

Therefore, the required term is $x^{22}$ .

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3 years ago