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3 years ago

11

Please help me!!!! :)

1 answer:

frutty [35]3 years ago

3 0

DE is the long narrow side, so judging from the picture, we can assume that DE is 6.

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You have to pay $32.75 for an item after a 25% markup. What was the original price?

MariettaO [177]

32.75 x .25
subtract the answer given with the original price, 32.75.
$24.56

3 0

3 years ago

Read 2 more answers

Is 15,36,39 a right triangle

NeTakaya

Answer:

Yes, is a right triangle

Step-by-step explanation:

To know if you can build a right triangle with those sides we have to make the following equality

h² = l1² + l2²

The hypotenuse is always the longest side so it should be 39

we check the equality of the equation

39² = 15² + 36²

1521 = 225 + 1296

1521 = 1521

Equality was fulfilled so it is a right triangle

7 0

3 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

A rectangular prism with length width and height has a volume of l w h what is the volume of a prism which has a base of 5 m by

Nady [450]

$v = l \: w \: h$
$v = 5 \times 3 \times 4$
$v = 15 \times 4$
$v = {60m}^{3}$

4 0

3 years ago

1/4 + 1/3= {Khan Academy btw}

zmey [24]

Answer: $\frac{7}{12}$

Step-by-step explanation:

$\frac{1}{4} +\frac{1}{3}=\frac{(1)(3)+(4)(1)}{(4)(3)} =\frac{3+4}{12} =\frac{7}{12}$

5 0

3 years ago