All categories

3 years ago

Sin idkxodnjxrubxhd hbdy budbhzbehzndhwxurbx

Mathematics

1 answer:

Sveta_85 [38]3 years ago

7 0

Answer:

That's the only thing that came up for Sin idkxodnjxrubxhd hbdy budbhzbehzndhwxurbx.

You might be interested in

In your lab, a substance's temperature has been observed to follow the function T(x) = (x − 4)3 + 6. The turning point of the gr

Montano1993 [528]

Yo sup??

To find the turning point we have to differentiate the given function.

T(x)=(x-4)3+6

T'(x)=3

Therefore the turning point is at 3

Hope this helps.

8 0

2 years ago

Decompose and use the distributive property to find the product of 5 × 7.3.

swat32

Answer:

36.5

Step-by-step explanation:

5(1+6.3)

6 0

2 years ago

Please help me with this

Cerrena [4.2K]

Answer:

30cm³

Step-by-step explanation:

volume of cuboid =L x B x H

=5 x 3 x 2

=30cm³

<h2><em>OladipoSeun</em><em>♡˖꒰ᵕ༚ᵕ⑅꒱</em></h2>

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

If P is the incenter of JKL, find each measure

d1i1m1o1n [39]

Answer:

Its really simple look it up at some site

6 0

2 years ago