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Inga [223]
3 years ago
11

Please find the areas of the four rectangles and ADD them together. Fill in the spaces below.

Mathematics
1 answer:
Andrei [34K]3 years ago
4 0
Total area = 9 x5 =45
Total area = (6+3)(4+1)=24+6+12+3=45
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Sebastian's pool is filled with 270 cubic meters of water. The base of the pool is 15 m long and 9 m wide. What is the height of
vodomira [7]
I got 2 m as the answer. 
4 0
3 years ago
If cos x=8/17, and 270° &lt; x &lt; 360°<br> what is cos(2x)?
svp [43]

Answer:  \bold{-\dfrac{161}{289}}

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

Given: cos x = \dfrac{8}{17}, x is in Quadrant 4

Use Pythagorean Theorem to find sin x:

8² + y² = 17²    

      y² = 17² - 8²

      y² = 289 - 64

      y² = 225

      y = 15

→    sin x = -\dfrac{15}{17}

Use the double angle formula to find cos (2x):

cos (2x) = cos^2 x - sin^2 x\\\\.\qquad=\bigg(\dfrac{8}{17}\bigg)^2-\bigg(-\dfrac{15}{17}\bigg)^2\\\\\\.\qquad=\dfrac{64}{289}-\dfrac{225}{289}\\\\\\.\qquad=-\dfrac{161}{289}

             

8 0
3 years ago
Read 2 more answers
I need help with #11
Troyanec [42]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\\\&#10;&#10;\begin{array}{rllll} &#10;% left side templates&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}&#10;\end{array}

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}&#10;\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

\bf \begin{array}{llll}&#10;&#10;&#10;\bullet \textit{ vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{{{  B}}}&#10;\end{array}

now, with that template in mind, let's take a peek at this function

\bf \begin{array}{lllcclll}&#10;y=&2(&1x&-2)^2&-4\\&#10;&\uparrow &\uparrow &\uparrow &\uparrow \\&#10;&A&B&C&D&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\&#10;\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\&#10;D=-4\impliedby \textit{vertical down shift of 4 units}

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically
8 0
3 years ago
Read 2 more answers
Integral of 17/(x^3-125)
daser333 [38]

Answer:

17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

Step-by-step explanation:

∫ 17 / (x³ − 125) dx

= 17 ∫ 1 / (x³ − 125) dx

= 17 ∫ 1 / ((x − 5) (x² + 5x + 25)) dx

Use partial fraction decomposition:

= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx

Use common denominator to find the missing coefficients.

A (x² + 5x + 25) + (Bx + C) (x − 5) = 1

Ax² + 5Ax + 25A + Bx² − 5Bx + Cx − 5C = 1

(A + B) x² + (5A − 5B + C) x + 25A − 5C = 1

Match the coefficients and solve the system of equations.

A + B = 0

5A − 5B + C = 0

25A − 5C = 1

A = 1/75

B = -1/75

C = -2/15

So the integral is:

= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx

Simplify:

= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx

Factor ½ from the numerator of the second fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx

Split the fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx

Multiply the last fraction by 4/4:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx

Complete the square:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx

Split the integral:

= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx

The first integral is:

∫ 1 / (x − 5) dx = ln│x − 5│

The second integral is:

∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)

The third integral is:

∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)

Plug in:

= 17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

4 0
3 years ago
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Ahat [919]
D.) 2 is the answer

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