Some one help me please

Pie

If you have a calculator:
1) To find 1%, divide by 100 (70 / 100 = 0.7) 

2) Then, multiply by the % you need (0.7 x 13 = 9.1) 

Therefore 70 - 9.1 = 60.9 

If you don't have a calculator: 
1) Find 50%, 10%, 5%, and 1%. With these, you can make any % you need. 
50% = X / 2 (= 35) 
10% = X / 10 (= 7) 
5% = first, do 10%, then divide that by 2 (= 3.5) 
1% = X / 100 (= 0.7) 
Where X is the number you want to work out % for 

2) Add those % up to get the amount you want 
10% + 1% + 1% +1% = 13% 
7 + 0.7 + 0.7 + 0.7 = 9.1 

Therefore 70 - 9.1 = 60.9
Hope that helps :)

7 0

3 years ago

Kylie has lemon and watermelon in a ratio of 10:71. How many watermelon does she have if she has 20 lemons

kkurt [141]

Answer:

She will have 142 watermelons

Step-by-step explanation:

First let's put our ratio

10 : 71

Since our lemons are now 20 instead of 10 we will divide 20 by 10 to see how many time we multiplied it by.

$20 \div 10$

which equals 2

To find our water melon we will multiply it by 2

$71 \times 2$

which equals 142

She will have 142 watermelons

7 0

3 years ago

The region bounded by y = e−x2, y = 0, x = 0, and x = b (b > 0) is revolved about the y-axis. (Round your answers to three de

kari74 [83]

Answer:

0.982 * $\pi$ = 3.085 cubic units

Step-by-step explanation:

To find the volume of the solid generated by a region that is bounded by lines where x = 0 , x (b) = 2 we apply the formula given in the attachment below

the required region boundaries are

y = e^-x2 , y = 0 , x = 0, x = b ( b >0)

attached below is a detailed solution of the problem

6 0

3 years ago

3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it

Juliette [100K]

(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)

(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...

... x ≈ 3.0473167

(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...

... x ≈ 3.2291234

(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...

... x ≈ 3.0477377

(e) Using Mathematica, the roots are found to be as shown in the second attachment. The highest root is about ...

... x ≈ 3.0466805180

_____

Comment on these methods

Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.

The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.

We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).