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2 years ago

You deposit $300 in a savings account. The account earns 2% simple interest per year.

Mathematics

1 answer:

Sholpan [36]2 years ago

8 0

Answer:

Answer:

a. ----> $10

b. ----> $110

Step-by-step explanation:

principle ( p ) = $100

time ( t ) = 10 years

rate ( r ) = 11%

simple interest = (p × r × t)÷ 100

= ( $100 × 11 × 10 )÷ 100

= 11000 ÷ 100

= $110

interest = simple interest - principle

interest = $110 - $100

= $10

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Which of the following is not a requirement of a standard form equation Ax+By=C? a)A ≥ 0. b)B ≥ 0. c)A and B are not both 0. d)A

Ghella [55]

<h2>Answer:</h2>

The following which is not a requirement of a standard form of equation Ax+By=C is:

b) B ≥ 0

<h2>Step-by-step explanation:</h2>

We know that the standard equation of a line is given by:

$Ax+By=C$

where A,B and C are integers and A is taken to be a non-negative integer i.e. (A≥0) also the greatest common factor of A,B and C is: 1.

Also A and B can't be both zero.

Hence, the correct option is:

Option: b

7 0

2 years ago

Read 2 more answers

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$