I will give brainiest to who ever can copy and answer this math problem first : 12342345791 + 3214567349=

A sector with a central angle measure of 7pi/4 (in radians) has a radius of 16 cm

strojnjashka [21]

Answer:

Area of sector = 704 cm ²

Step-by-step explanation:

Given:

Central angle = 7pi/4 (in radians)

Radius = 16 cm

Find:

Area of sector = ?

Computation:

Central angle = $\frac{7}{4}\pi \times \frac{180}{\pi } =315$

$Area\ of \ sector=\frac{\theta}{360}\pi r^2\\\\Area\ of \ sector=\frac{315}{360}\pi 16^2\\\\Area\ of \ sector=704 cm^2$

Area of sector = 704 cm ²

5 0

3 years ago

Still don’t get it idk lol

KonstantinChe [14]

Answer:

XQ = 12

Step-by-step explanation:

In this question it asks for XQ instead of x, but you need to find x first.

1. Find x.

XQ is half of MQ so this statement is true: 3x - 3 = 2(2x - 6)

Solve for x.

3x - 3 = 2(2x - 6)

Distribute 2.

3x - 3 = 4x - 12

Isolate x.

3x = 4x - 9

-x = -9

Divide -1 out.

x = 9

Now solve for XQ, which the equation is given.

XQ = 2(9) - 6

XQ = 18 - 6

XQ = 12

6 0

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

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

5 0

2 years ago

Select the table that represents the function x2 + 4x

Anvisha [2.4K]

Answer:

Table 1

Step-by-step explanation:

Check Table 1:-

x = -4 gives y = (-4)^2 + 4(-4) = 0

x = -2 gives y = (-2)^2 - 2(-2) = -4

x = 0 gives 0 + 0 = 0

x = 1 gives 1^2 + 4 = 5

That's the one!

4 0

3 years ago

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100%

Pavlova-9 [17]

The answer is E.$1,170

7 0

3 years ago

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