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NeX [460]
3 years ago
15

What is the linear factorization of the function F(x) = 5x^3 + 5x ?

Mathematics
1 answer:
Aliun [14]3 years ago
5 0
5x^3 + 5x = 5x(x^2 + 1)
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The sign shows distances from a rest stop to the exits for different towns along a straight section of highway. The state depart
torisob [31]
Refer to the diagram shown below.

The exit for Freestone is built midway between Roseville and Edgewood,
therefore the distance from O to the new exit is
(1/2)*(33+55) = 44 mi.

Let x =  distance from Midtown to the new exit.
Because the distance from O to the new exit is equal to (x + 17), therefore
x + 17 = 44
x = 44 - 17 = 27 mi.

Answer:
When the new exit is built, the distance from the exit for Midtown to the exit for Freestone will be 27 miles.

3 0
3 years ago
21.58 correct to 2 decimal places
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Answer:

Step-by-step explanation:

its 22.00 or just 22

3 0
3 years ago
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Generate two numerical patterns
Klio2033 [76]

Answer:

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Step-by-step explanation:

2,4,6,8,10

Pattern Rule:Add 2

1,2,4,8,16

Pattern:Multiply by 2

Sorry these were simple.

8 0
3 years ago
Simplify:
liubo4ka [24]

Answer:

78.5÷3.14=25

56.272×41.2=2318.4064

429×338×712=103241424

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Step-by-step explanation:

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8 0
3 years ago
1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.
m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = \frac{9}{7+9}\times (x)

                                      = \frac{9}{16}\times (x)

And the weight of zinc = \frac{7}{7+9}\times (x)

                                      = \frac{7}{16}\times (x)

If the weight of zinc = 31.5 kg

31.5 = \frac{7}{16}\times (x)

x = \frac{16\times 31.5}{7}

x = 72 kgs

Therefore, weight of copper = \frac{9}{16}\times (72)

                                               = 40.5 kgs

2). i). 2 : 3 = \frac{2}{3}

        4 : 5 = \frac{4}{5}

Now we will equalize the denominators of each fraction to compare the ratios.

\frac{2}{3}\times \frac{5}{5} = \frac{10}{15}

\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}

Since, \frac{12}{15}>\frac{10}{15}

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = \frac{11}{19}

    19 : 21 = \frac{19}{21}

By equalizing denominators of the given fractions,

\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}

And \frac{19}{21}\times \frac{19}{19}=\frac{361}{399}

Since, \frac{361}{399}>\frac{231}{399}

Therefore, 19 : 21 > 11 : 19

iii). \frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}

             =\frac{3}{2}

     \frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}

              = \frac{4}{3}

Now we equalize the denominators of the fractions,

\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}

And \frac{4}{3}\times \frac{2}{2}=\frac{8}{6}

Since \frac{9}{6}>\frac{8}{6}

Therefore, \frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4} will be the answer.

IV). 1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}

                  =\frac{6}{5}\times \frac{3}{4}

                  =\frac{18}{20}

                  =\frac{9}{10}

Similarly, \frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}

                       =\frac{4}{15}                  

By equalizing the denominators,

\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}

Similarly, \frac{4}{15}\times \frac{20}{20}=\frac{80}{300}

Since \frac{270}{300}>\frac{80}{300}

Therefore, 1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}

V). If a : b = 6 : 5

     \frac{a}{b}=\frac{6}{5}

        =\frac{6}{5}\times \frac{2}{2}

        =\frac{12}{10}

  And b : c = 10 : 9

  \frac{b}{c}=\frac{10}{9}

 Since a : b = 12 : 10

 And b : c = 10 : 9

 Since b = 10 is common in both the ratios,

 Therefore, combined form of the ratios will be,

 a : b : c = 12 : 10 : 9

7 0
3 years ago
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