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3 years ago

HELP ASAPP!! Distribute: 3(-2x + 9)

Mathematics

2 answers:

Dima020 [189]3 years ago

5 0

Answer:

-6x+27

hope this helps

have a good day :)

Step-by-step explanation:

Oksana_A [137]3 years ago

4 0

Answer:

-6x+27

Step-by-step explanation:

3(-2x + 9)

-6x+27

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Given the following, find the value of x.<br> (4x+45)<br> (5x-18)

VladimirAG [237]

Step-by-step explanation:

Hey ..but there is nothing from which question can be equated ..

like if they are equal to zero..

(4x+45)=(5x-18)

i am giving answer for both scenarios

(4x+45)=0

4x=-45

x=-45/4

x=(-11.25)

(5x-18)=0

5x=18

x=18/5

x=3.6

Incase if they both are equal to each other...

(4x+45)=(5x-18)

(45+18)=(5x-4x)

63=x

x=63

Hope it helps you

Have a nice day

8 0

3 years ago

Read 2 more answers

After a long study, tree scientists conclude that a eucalyptus tree will grow at the rate of 0.5 6/ (t+4)3 feet per year, where

kipiarov [429]

Answer:

Step-by-step explanation:

Given the rate at which an eucalyptus tree will grow modelled by the equation 0.5+6/(t+4)³ feet per year, where t is the time (in years).

The amount of growth can be gotten by integrating the given rate equation as shown;

$\int\limits {0.5 + \frac{6}{(t+4)^{3} } } \, dt \\= \int\limits {0.5} \, dt + \int\limits\frac{6}{(t+4)^{3} } } \, dx } \, \\= 0.5t +\int\limits {6u^{-3} } \, du \ where \ u = t+4 \ and\ du = dt\\= 0.5t + 6*\frac{u^{-2} }{-2} + C\\= 0.5t-3u^{-2} +C\\= 0.5t-3(t+4)^{-2} + C$

a) The number of feet that the tree will grow in the second year can be gotten by taking the limit of the integral from t =1 to t = 2

$\int\limits^2_1 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^2_1\\= [0.5(2)-3(2+4)^{-2}] - [0.5(1)-3(1+4)^{-2}]\\= [1-3(6)^{-2}] - [0.5-3(5)^{-2}]\\ = [1-\frac{1}{12}] - [0.5-\frac{3}{25} ]\\= \frac{11}{12}-\frac{1}{2}+\frac{3}{25}\\ = 0.9167 - 0.5 + 0.12\\= 0.5367feet$

b) The number of feet that the tree will grow in the third year can be gotten by taking the limit of the integral from t =2 to t = 3

$\int\limits^3_2 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^3_2\\= [0.5(3)-3(3+4)^{-2}] - [0.5(2)-3(2+4)^{-2}]\\= [1.5-3(7)^{-2}] - [1-3(6)^{-2}]\\ = [1.5-\frac{3}{49}] - [1-\frac{1}{12} ]\\ = 1.439 - 0.9167\\= 0.5223feet$

c) The total number of feet grown during the second year can be gotten by substituting the value of limit from t = 0 to t = 2 into the equation as shown

$\int\limits^2_0 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^2_0\\= [0.5(2)-3(2+4)^{-2}] - [0.5(0)-3(0+4)^{-2}]\\= [1-3(6)^{-2}] - [0-3(4)^{-2}]\\ = [1-\frac{1}{12}] - [-\frac{3}{16} ]\\= \frac{11}{12}+\frac{3}{16}\\ = 0.9167 - 0.1875\\= 0.7292feet$