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3 years ago

One fourth plus ten sixteenths

Mathematics

2 answers:

Anettt [7]3 years ago

7 0

Answer:

fourteen-sixteenths

Step-by-step explanation:

Alex73 [517]3 years ago

6 0

Answer:

7/ 8

(Decimal: 0.875)

Step-by-step explanation:

1/4+10/16

1/4+5/8 (simplfy into simpliest terms)

2/8+5/8 (find the common denominator)

2+5/8 (add the numerators)

Answer=7/8

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120% of 60 with a small explanation

Shtirlitz [24]

120(60)

120 ÷ 100 = 1.2
120 / 100 = 1.2

Take off the "60" and multiply it by 1.2

1.2(60) = 72, right ?

So, therefore your answer would have to be : 72

8 0

4 years ago

A board 120 inches long is divided into 2 sections. If the ratio of the 2 sections is 3:5, what are the lengths of the sections?

riadik2000 [5.3K]

Answer: D.45 inches, 75 inches

Step-by-step explanation:

Hi, to answer this question, first, we have to divide the board's length by the number of parts that conforms the 2 sections.

Since the ratio is 3/5, one section is made of 3 parts and the other section is made of 5 parts. 3+5 = 8 parts

120 / 8 = 15 inches

Each part measures 15 inches.

Now, we have to multiply each part's length by the number of parts that each section has. (3 and 5)

15x 3 = 45 inches

15 x 5 = 75 inches

3 1

4 years ago

If you can buy 1/4 pizza for 5 dollars, how much can you purchase for 8 dollars?

Fynjy0 [20]

Answer: 2/5 pizza

Step-by-step explanation:

8 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

If tony made 378$ for 18 hours of work how many hours would he work to make 231$?

Hatshy [7]

Answer:

Step-by-step explanation:

378 / 18 = 21 // Money p/h

231 / 21 = 11

5 0

3 years ago

Read 2 more answers