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zhenek [66]
3 years ago
14

The sum of the lengths of any two sides of a triangle must be less than the length of the remaining side. True or false?

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
6 0

False. The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.

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Help please!!!!!!!!!!!
natita [175]
3x-12 ≥7x+4
Subtract 3x from both sides
-12 ≥4x+4
Subtract 4 from both sides
-16 ≥4x
Divide each side by 4
-4 ≥x
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6 0
3 years ago
Mary ravi and Alan have a total of $62 in their wallets. Mary has $6 more than Ravi. Alan has 2 times what Mary has. How much do
mixas84 [53]

Answer:

Marry = $17

Ravi = $11

Alan = $34

Step-by-step explanation:

Let us assume

Mary be X

Ravi be Y

Alan be Z

So, the equation in total is

X + Y + Z = 62

It is given that

X = Y + 6

Z = 2X

Now we put the Z value in the total equation, so it would be

X + Y + 2X = 62

3X + Y = 62  .................................... (1)

And can we write

X = Y + 6

X - Y = 6 ................................. (2)

Now substitution these two equations,

3X + Y = 62

X - Y = 6

2X = 68

X = $17

So, Z = 2 × 17 = $34

And, Y

$17 + Y + $34 = $62

So, Y = $11

5 0
2 years ago
Which of the following is a valid probability distribution? A 2-column table labeled Probability Distribution A has 4 rows. The
Ksenya-84 [330]

Answer:

i think its D on edg

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
3 years ago
Identify the coefficient in the term 7x^2y^3
Advocard [28]

Answer:

<h3>LARGEST COEFFICIENT = 7</h3><h3>SMALLEST COEFFICIENT=2</h3><h3>STANDARD FORM=7,0,0,0,3,2,0,0</h3><h3>COEFFICIENTS =7,3,2</h3>
3 0
2 years ago
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