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3 years ago

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Mathematics

1 answer:

Bad White [126]3 years ago

6 0

Answer:

option 2 : $=x^4 - 3x^3 +x^2 -4$

Step-by-step explanation:

Given P(x) = R(x) - C(x)

$R(x) = 2x^4 - 3x^3 + 2x - 1 \ , \ C(x) = x^4 - x^2 + 2x + 3$

$P(x) = (2x^4 - 3x^3 + 2x -1) - (x^4 -x^2 + 2x +3)$

$=2x^4 -3x^3 + 2x -1 -x^4 +x^2 -2x -3$

$= (2x^4 - x^4) - 3x^3 +x^2 +(2x -2x) + (-1 -3)$ $[arranging \ terms]$

$=x^4 - 3x^3 +x^2 -4$

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Paul, colin and brian are waiters.One night the restaurant earns tips totalling £78.40.

riadik2000 [5.3K]

Answer:

£39.20

Step-by-step explanation:

2+1+5=8

£78.40/8 = 9.8

9.8*5= 49

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3 years ago

For the right triangle shown, the lengths of two sides are given. find the third side. leave your answer in simplified, radical

lisov135 [29]

If c is the hypotenuse and b is the leg, then the third side (a):
$a= \sqrt{c^2-b^2}= \sqrt{16^2-9^2}= \sqrt{256-81}= \sqrt{175}=5 \sqrt{7}$

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$a= \sqrt{b^2+c^2}= \sqrt{9^2+16^2}= \sqrt{81+256}= \sqrt{337}$

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3 years ago

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Step-by-step explanation:

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3 years ago

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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

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

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2 years ago

Margaret [11]

Complete Question: In Rhombus QRST, diagonals RT and QS are drawn. If m<TSR = 104°, then which of the following is the measure of <SRT?

*the diagram of the rhombus is in the attachment below

Answer:

m<SRT = 38°

Step-by-step explanation:

A rhombus has 4 equal sides. Also, diagonals of a rhombus bisect each other Therefore:

∆RST form the rhombus given is an isosceles triangle with two equal lengths, ST and SR, and two equal base angles, <T and <R.

Given that <TSR = 104°, thus,

m<SRT = (180 - 104)/2 = 76/2

m<SRT = 38°

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3 years ago