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Tamiku [17]
3 years ago
7

Simplify 8 3{x - 2[x 5(x 3)]}.

Mathematics
1 answer:
Anna11 [10]3 years ago
5 0
8 3{x - 2[x 5(x 3)]}i dont know if it is 8^3 or 83 or 8x3 but it is the last to multiplied then i can just adjust the answer acordingly
first muliply the most inner term8 3{x - 2[x 5(x 3)]}= 8 3{x - 2[x 8]}8 3{x 9- 2x 8}
if it is 8^3 = 512
8 3{x 9- 2x 8}= 512{x 9- 2x 8}= 512x 9- 1024x 8}
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D+7/−3=4<br> What is d?<br> Thanks!
givi [52]

Answer

\boxed {d= \frac{19}{3}}


Solution

Simplify both sides of the equation.


\frac{7}{3} + \frac{-7}{3}


We are left with d = 4+\frac{7}{3}


Add

\frac{12}{3}+\frac{7}{3} =\frac{19}{3}


Hence, d = \frac{19}{3}


7 0
3 years ago
What is usually the best way to display data if you have more than 10 results?
mestny [16]

Answer:

Probably on a pie chart.

Step-by-step explanation:

You can color code each piece of data, and show the percentage for each on the pie chart.

5 0
3 years ago
The purchase price of a camcorder is $640. What is the total price if the sales tax rate is 8.5%?
iVinArrow [24]
100%-8.5%=91.5%
After tax, the price is 91.5% of original price.
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8 0
3 years ago
Read 2 more answers
Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4
lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

                       f(x) =  { x^2+3x+4    , x < 3

                       f(x) =  { x^2+3x+4     , x≥3

                        Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:      

                       f(x) = x^2+3x+4

- The area "A1" bounded by f(x) is given as:

                      A1 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { -3 , 3 ] = [ a , b ]:

                     A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

                     f(x) = 4x+10

- The area "A2" bounded by f(x) is given as:

                      A2 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { 3 , 4 ] = [ a , b ]:

                     A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

                     A = A1 + A2

                     A = 42 + 26

                     A = 68 unit^2

8 0
2 years ago
PLSS HELPP DUE IN 20 MIN. I NEED SOME STEP BY STEP TOO.
Veseljchak [2.6K]

The length of GH is 15.

<h3>How to calculate the length?</h3>

GH = IG

GH = 5x - 10

IG = 3x

5x - 10 = 3x

5x - 3x = 10

2x = 10

x = 10/2 = 5

GH = 5x - 10

= 5 * 5 - 10 = 25 - 10

GH = 15

<ABD = < DBC

<ABD = 10y°

<DBC = (8y + 4)°

10y° = (8y + 4)°

10y = 8y + 4

10y - 8y = 4

2y = 4

y = 4/2 = 2

The value of y is 2

<DBE = <DBC + <CBE

<DBC = 10y°

<CBE = <DBC = 10y°

y = 2

<DBC = <CBE = 10y° = 10*2° = 20°

<DBE = 20° + 20° = 40°

Learn more about angles on:

brainly.com/question/25716982

#SPJ1

8 0
1 year ago
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