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3 years ago

John has $65 in his piggy bank. He wants to buy a new gaming

Mathematics

1 answer:

docker41 [41]3 years ago

6 0

Yea it will take him 16 days

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Simplify each expression 9 to the second power

olga_2 [115]

9 to the 2nd power would = 81.

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3 years ago

What is the slope of the line?<br> Pls help! I’ll mark the brainliest

VLD [36.1K]

Answer:

m = -3

Step-by-step explanation:

Two poin on this line are (-2, 3) and (0, -3). Both points are formed where grid lines meet precisely.

The change in x as we go from (-2, 3) to (0, -3) is +2 and the change in y is -6. Thus, the slope of the line shown is m = rise/run = -6/2, or m = -3.

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3 years ago

hello hello the area of a triangle is 32 square in the height of the triangle is 8 in what is the length of the base of a triang

solniwko [45]

As we know
area of triangle = 1/2 *height*base
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32=4*base
8 =base

3 0

3 years ago

Read 2 more answers

Simplify the following expression 4w x 3w x 2y x 2y

Anestetic [448]

Answer:

48w^2y^2

I used a simplify calculator I hope this is correct.

4 0

3 years ago

Read 2 more answers

<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