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3 years ago

The decimal number 11.8 can be read as "eleven and eight tens." True or False

Mathematics

2 answers:

melamori03 [73]3 years ago

4 0

Answer is False
11.8 should be read as "eleven and eight tenths."

Semmy [17]3 years ago

3 0

I mean it could be read that way but that's the way most people read fractions. 11 and 8/10. Most people I know read decimals like that as 11 point 8. So yes I suppose it's true; it's not wrong, but most people don't say it like that. And I don't know how they taught you to read it in your math course.

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There Is 2 cars and 6 seats in each car how many people go in each car and do we in ignore the remainder or we don't

KIM [24]

6 people go in each car because since there is only six seats 6x1 equal 6 and there i no remainder

5 0

3 years ago

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The equation V = 16300 (0.94)^t represents the value (in dollars) of a car t years after its purchase. Use this equation to comp

Dafna1 [17]

Solution:

Given:

$V=16300(0.94)^t$

The value of a car after t - years will depreciate.

Hence, the equation given represents the value after depreciation over t-years.

To get the rate, we compare the equation with the depreciation formula.

$\begin{gathered} A=P(1-r)^t \\ \text{where;} \\ P\text{ is the original value} \\ r\text{ is the rate} \\ t\text{ is the time } \end{gathered}$

Hence,

$\begin{gathered} V=16300(0.94)^t \\ A=P(1-r)^t \\ \\ \text{Comparing both equations,} \\ P=16300 \\ 1-r=0.94 \\ 1-0.94=r \\ r=0.06 \\ To\text{ percentage,} \\ r=0.06\times100=6\text{ \%} \\ \\ \text{Hence, } \\ P\text{ is the purchase price} \\ r\text{ is the rate} \end{gathered}$

Therefore, the value of this car is decreasing at a rate of 6%. The purchase price of the car was $16,300.

5 0

1 year ago

What is the first step in solving the equation 3.5 n + 6.4 = 42.5?

miss Akunina [59]

I would add like terms, meaning subtract 6.4 from both sides.

3.5n = 36.1

7 0

3 years ago

Read 2 more answers

Find the sum of 12a + 7b, −6a − 9, and 14a − 12b.

Ivan

The answer is: [A]: " 20a − 5b − 9 " .
______________________________________
Explanation:
_______________________________________________
(12a + 7b) + (−6a − 9) + (14a − 12b) =

12a + 7b + 1(−6a − 9) + 1(14a − 12b) =

12a + 7b + (1*-6a) + (1*-9) + (1*14a) + (1* -12b) =

12a + 7b − 6a − 9 + 14a − 12b = ?

Combine the "like terms:

12a − 6a + 14a = 20a ;

7b − 12b = - 5b ;

and then we have "-9" ;
_____________________________________________________
So, write as: " 20a − 5b − 9 " ; which is: Answer choice: [A].
______________________________________________________

8 0

3 years ago

Read 2 more answers

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

3 years ago