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3 years ago

6

Grace sold 5/8 of her stamp collection what is this amount as a decimal

2 answers:

Alex_Xolod [135]3 years ago

7 0

The answer for your questions is 0.625

fenix001 [56]3 years ago

6 0

0.625
divide 5 by 8 is 0.625 rounded would be 0.63

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Which of the following is the value of x?<br> A:2<br> B:11<br> C:20<br> D29

AleksAgata [21]

Answer:

C. 20

Step-by-step explanation:

45 + 5x + 35 = 180

5x + 80 = 180

subtract 80 from both sides

5x = 100

divide by 5

x = 20

3 0

3 years ago

Read 2 more answers

What is the value of y?<br><br> A. <br> 16<br><br> B. <br> 48<br><br> C. <br> 64<br><br> D.<br> 132

lorasvet [3.4K]

Answer:

3x+3x+7x+20+7x+20=360

20x=360-20

20x=320

x=16

therefore x=16 so,

7x+20+y=180

but x=16

7(16)+20+y=180

y=180-132

y=48

6 0

2 years ago

19. Find the perimeter of a triangle if the side lengths are 3 1/2 ft, 5 2/3 ft,<br> and 4 1/6 ft.

vagabundo [1.1K]

Answer: 13 1/3

Step 1: Find the LCD

Step 2: Add the fractions

Step 3: Simplify

3 0

3 years ago

The sum of the first n terms of a geometric series is 364? The sum of their reciprocals 364/243. If the first term is 1, find n

Afina-wow [57]

If the geometric series has first term $a$ and common ratio $r$ , then its $N$ -th partial sum is

$\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}$

Multiply both sides by $r$ , then subtract $rS_N$ from $S_N$ to eliminate all the middle terms and solve for $S_N$ :

$rS_N = ar + ar^2 + ar^3 + \cdots + ar^N$

$\implies (1 - r) S_N = a - ar^N$

$\implies S_N = \dfrac{a(1-r^N)}{1-r}$

The $N$ -th partial sum for the series of reciprocal terms (denoted by $S'_N$ ) can be computed similarly:

$\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}$

$\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}$

$\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}$

$\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}$

We're given that $a=1$ , and the sum of the first $n$ terms of the series is

$S_n = \dfrac{1-r^n}{1-r} = 364$

and the sum of their reciprocals is

$S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}$

By substitution,

$\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243$

Manipulating the $S_n$ equation gives

$\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363$

so that substituting again yields

$r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}$

and it follows that

$r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}$

5 0

2 years ago

Steps for how to use a number line to multiply a 2-digit number by 20

Feliz [49]

its is the same as multiplying normally .

For example 1-2-3-4-5-6-7-8- 2 times 2

you just take the starting number (2) and times it by the other number (2) and youll get 4

6 0

3 years ago