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3 years ago

Find the length of x in simplest radical form with a rational denominator

Mathematics

2 answers:

Alisiya [41]3 years ago

8 0

Answer:

Step-by-step explanation:

$\cos(30^o)=\dfrac{\sqrt3}2$

From triangle: $\cos(30^o)=\dfrac x2$

Therefore:

$\dfrac x2=\dfrac{\sqrt3}{2}\\\\x=\sqrt3$

aleksandrvk [35]3 years ago

7 0

It is equals to 3 I think

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Verify that the intermediate value theorem applies to the indicated interval and find the value of c guaranteed by the theorem.

Inessa05 [86]

The intermediate value theorem applies to the indicated interval and the importance of c guaranteed by the theorem is c=2,3.

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2-1

and interval = [4]

since, function fext is continuous in ginen

interval. And also

+(4) = 42+4

4-1

20 = 6667

$(5/4) = ($145/2

stone-1

= 5.833

simle, f(4) > $(5/2), hence Intermediate

Theorem & applies to the indicated

proved.

Now,

= 6

C-1

C-5c +6 = 0

C=2 or c=3

3 or

C= 2, 3

<= 2

Learn more about theorem here brainly.com/question/26594685

#SPJ4

8 0

2 years ago

A 12-ounce bottle of shampoo costs $2.79. A 16-ounce bottle of shampoo costs $3.25. Which statement is true?

JulsSmile [24]

That would be C! Good Luck!

6 0

3 years ago

Answer this if you can 37÷264

docker41 [41]

.1401515152 is not the answer

6 0

3 years ago

Read 2 more answers

Plz help ill give u brainlist pla asap

zhenek [66]

Area of base = 1/2 x 18 x 15.6 = 140.4
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7 0

3 years ago

There are given complex numbers

Karolina [17]

Ok... so, we know that both pairs are conjugate of each other,
so.. both are complex solutions, namely, both are a+bi type

what's "a", we dunno, what is "b", we dunno either
but for z1, we know that a = m-2n-1 and b = 5m-4n-6
for z2, a = 3m-n-6 and b=m-5n-3

for those two folks to be conjugate of each other, they'd be
(a+bi)(a-bi)
notice, the "a"s are the same on z1 as well as z2
the "b" are the same also BUT the z2 "b" is negative,
in order to be a conjugate, regardless of that, the "b"s are the same

so, whatever (m-2n-1) is, is the same as <span>(3m-n-6), since a = a
and whatever </span><span>(5m-4n-6) is, will be the same value, since b =b,
the "a" and "b" in each conjugate, is the same value, thus one can say
</span> $\begin{array}{cccll}
z1=&(m-2n-1)+i&(5m-4n-6)\\
&\uparrow &\uparrow \\
&a&b\\
&\downarrow &\downarrow \\
z2=&(3m-n-6)+i&(m-5n-3)
\end{array}
\\ \quad \\

\begin{cases}
m-2n-1=3m-n-6\implies 0=2m+n-5\\
5m-4n-6=m-5n-3\implies 4m+n-3=0
\end{cases}
\\ \quad \\
\textit{now, solving that by using elimination}$
<span>
</span> $\begin{array}{llcll}
2m+n-5=0&\leftarrow \times -2\implies &-4m-2n+10=0\\
4m+n-3=0&&4m+n-3=0\\
&&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
&&-n+7=0
\end{array}
\\ \quad \\
or\qquad 7=n$
<span>
and pretty sure you can find "m" from there
once you have both, substitute in z1 or z2, to get the values for "a" and "b"</span>

7 0

3 years ago