All categories

2 years ago

Please help and please show work!!!

Mathematics

1 answer:

mina [271]2 years ago

3 0

9514 1404 393

Answer:

a. 61

Step-by-step explanation:

The remainder theorem tells you the remainder of f(x)/(x -3) is equal to f(3). Using x=3 in the expression, we have ...

4x³ -2x² -10x +1

= ((4x -2)x -10)x +1

= ((4·3 -2)(3) -10)(3) +1 = (10·3 -10)(3) +1 = 20·3 +1

= 61

The remainder from division by (x -3) is 61.

You might be interested in

What is the growth factor that corresponds to a product that increases its value first by 2%, and then increases by 5% of

BartSMP [9]

Answer:

1.19

Step-by-step explanation:

1+0.02+0.05+0.12 = 1.19

8 0

3 years ago

Write an equation for the line parallel to the given line that contains C.<br> Help

tangare [24]

There is no photo shown

6 0

3 years ago

How many roots does the equation (8/(x^2 - 16) )+ 1 = 1/(x -4) have?<br><br> Plz show ALL STEPS

balandron [24]

Answer:

Step-by-step explanation:

4 0

3 years ago

H(x)={(-7,-1),(0,0),(5,8)} domain and range

Bingel [31]

Answer:

The domain is -7, 0, and 5

The range is -1, 0, and 8

Step-by-step explanation:

The domain of a set of points is the x-value. In this case the x-values are -7, 0, and 5 respectively. The range of a set of points is the y-value so in this case the range is -1, 0, and 8.

7 0

2 years ago

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

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

8 0

3 years ago