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STatiana [176]
3 years ago
7

Explain how you know that in ?×6,273=6,273, the ? Will be 1

Mathematics
2 answers:
irakobra [83]3 years ago
8 0

Answer:

There would be 1.

Step-by-step explanation:

In real analysis,

1 is a multiplication identity,

Also, a real number after multiplying by the multiplicative identity gives the same number,

That is,

a . 1 = a = 1 . a ∀ a ∈ R

Since, 6273 is a real number,

Thus,

1 × 6273 = 6273

Hence, proved....

grigory [225]3 years ago
6 0
Anything more or less than 1 would result in a different outcome, and anything times 1 results in itself.
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yarga [219]

Answer:

Rewrite using the commutative property of multiplication.

− 4 t s

Step-by-step explanation:

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3 years ago
Please find the SURFACE AREA of this figure... Write the formula and plug in the values... Write your answer with units and box
sashaice [31]

Answer:

hope this helps you

6 0
3 years ago
Plsss help it’s due today in an hour!!! (test grade)
Rudik [331]

Answer:

the area is 50.24 cm^{2}.

Step-by-step explanation:

The formula to find area of a circle is A=πr^{2}. A is area, and r is for radius. Since we have the diameter here, we just need to divide by 2 to find the radius. 8÷2=4, so there, the radius is 4. Now we need to square 4, and do 4x4, which equals 16. Now to multiply 16 by π, or 3.14. 16 x 3.14 = 50.24. The final answer is 50.24 cm^{2}. Hope this helped! :]

8 0
3 years ago
Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it
Zanzabum

Answer:

\beta=45\degree\:\:or\:\:\beta=135\degree

Step-by-step explanation:

We want to solve \tan \beta \sec \beta=\sqrt{2}, where 0\le \beta \le360\degree.

We rewrite in terms of sine and cosine.

\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}

\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}

Use the Pythagorean identity: \cos^2\beta=1-\sin^2\beta.

\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}

\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)

\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta

\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0

This is a quadratic equation in \sin \beta.

By the quadratic formula, we have:

\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }

\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }

\sin \beta=\frac{2}{2\cdot \sqrt{2} } or \sin \beta=\frac{-4}{2\cdot \sqrt{2} }

\sin \beta=\frac{1}{\sqrt{2} } or \sin \beta=-\frac{2}{\sqrt{2} }

\sin \beta=\frac{\sqrt{2}}{2} or \sin \beta=-\sqrt{2}

When \sin \beta=\frac{\sqrt{2}}{2} , \beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )

\implies \beta=45\degree\:\:or\:\:\beta=135\degree on the interval 0\le \beta \le360\degree.

When  \sin \beta=-\sqrt{2}, \beta is not defined because -1\le \sin \beta \le1

4 0
3 years ago
Percent increase or decrease can be found using an equation? Write the equation below.
Nataly [62]
% increase = Increase ÷ Original Number × 100.<span>
<span>
</span></span>
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