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pashok25 [27]
2 years ago
13

If JKLM is a trapezoid, which statement must be true?

Mathematics
1 answer:
QveST [7]2 years ago
5 0

Answer:

1

Step-by-step explanation:

both are parrelel lines

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Which expression represents the value of the series below?<br> 3+ 7 + 11 + 15 + ... + 1,671
torisob [31]

Answer:

Step-by-step explanation:

This is an Arithmetic Series with common difference 4 and first term 3

so the nth term an = 3 + (n -1)4

= 4n - 1.

The sum of n terms

= n/2 (a1 + L)

= n/2(3 + 1671)

=  837n.

There are (1671-3) / 4 + 1 = 418 terms in the series,

so the total value of the series is 837*418

= 349,866.

.

4 0
2 years ago
What is the factored form of this expression? 27m^3 +125n^3
ddd [48]

Answer:

\left(3m+5n\right)\left(9m^2-15mn+25n^2\right)

Step-by-step explanation:

27m^3+125n^3=\left(3m\right)^3+\left(5n\right)^3

Apply sum of cubes formula: x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)

\left(3m\right)^3+\left(5n\right)^3=\left(3m+5n\right)\left(3^2m^2-3m\times 5n+5^2n^2\right)

\left(3m+5n\right)\left(9m^2-15mn+25n^2\right)

5 0
2 years ago
Read 2 more answers
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
Brayden is going to invest in an account paying an interest rate of 4.3% compounded
Julli [10]

Answer:

810

Step-by-step explanation:

4 0
3 years ago
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Sebastian has 12 more trophies than Megan. Megan has t trophies. How many trophies does Sebastian have? Write your answer as an
ikadub [295]

Answer:

t + 12 = s

Step-by-step explanation:

t (Megan's trophies) + 12 = s (Sebastian's trophies)

8 0
3 years ago
Read 2 more answers
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