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Citrus2011 [14]
3 years ago
8

A polynomial with three terms

Mathematics
1 answer:
NikAS [45]3 years ago
3 0
Its called a trinomial
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Triangle ABC has vertices with the following coordinates:
Triss [41]

Answer:

A'= (-4,2)

B'= (-3,3)

C'= (-2,0)

Step-by-step explanation:

3 0
2 years ago
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Margo can purchase tile at a store for $0.69 per tile and rent a tile saw for $36. At another store she can borrow the tile saw
TEA [102]

Answer:

<h3>30 tiles need to buy for same cost in both store.</h3>

Step-by-step explanation:

Let x be the number of tiles to buy for same cost in both store.

18+0.69 · n = 1.29 · n,

1.29n -0.69n =18 or

0.6n =18

n =18/0.6=30

Therefore, 30 tiles need to buy for same cost in both store.[Ans]

7 0
2 years ago
Read 2 more answers
85.
Zanzabum

Answer:

A. you need 2 1/2 ounces

B. you need 73,9 milliliters

Step-by-step explanation:

A

We have the ratio 1/2 ounces per gallon

And the mop bucket holds 5 gallons

Then we need to multiply 1/2*5

we got 2 1/2 ounces of desifectant per 5 gallons

B

1 ounces equal 29,57 milliliters

if we have 2 1/2 ounces then we multiply 2,5 *29,57

the answer is 73,9 milliliters

4 0
2 years ago
Complete the equation.<br> 2 x 4 =<br> X 2
rewona [7]

Answer:

4 is correct answer.

Step-by-step explanation:

That because it contains property of communicative. That is a×b=b×a.

4 0
1 year ago
Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))
Sonja [21]

Answer:

\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}

Let a=\sin^{-1}(\frac{2}{3}).

With some restriction on a this means:

\sin(a)=\frac{2}{3}

We need to find \tan(a).

\sin^2(a)+\cos^2(a)=1 is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

(\frac{2}{3})^2+\cos^2(a)=1

\frac{4}{9}+\cos^2(a)=1

Subtract 4/9 on both sides:

\cos^2(a)=\frac{5}{9}

Take the square root of both sides:

\cos(a)=\pm \sqrt{\frac{5}{9}}

\cos(a)=\pm \frac{\sqrt{5}}{3}

The cosine value is positive because a is a number between -\frac{\pi}{2} and \frac{\pi}{2} because that is the restriction on sine inverse.

So we have \cos(a)=\frac{\sqrt{5}}{3}.

This means that \tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}.

Multiplying numerator and denominator by 3 gives us:

\tan(a)=\frac{2}{\sqrt{5}}

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

\tan(a)=\frac{2\sqrt{5}}{5}

Let's continue on to letting b=\cos^{-1}(\frac{1}{7}).

Let's go ahead and say what the restrictions on b are.

b is a number in between 0 and \pi.

So anyways b=\cos^{-1}(\frac{1}{7}) implies \cos(b)=\frac{1}{7}.

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of b.

\cos^2(b)+\sin^2(b)=1

(\frac{1}{7})^2+\sin^2(b)=1

\frac{1}{49}+\sin^2(b)=1

Subtract 1/49 on both sides:

\sin^2(b)=\frac{48}{49}

Take the square root of both sides:

\sin(b)=\pm \sqrt{\frac{48}{49}

\sin(b)=\pm \frac{\sqrt{48}}{7}

\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}

\sin(b)=\pm \frac{4\sqrt{3}}{7}

So since b is a number between 0 and \pi, then sine of this value is positive.

This implies:

\sin(b)=\frac{4\sqrt{3}}{7}

So \tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}.

Multiplying both top and bottom by 7 gives:

\frac{4\sqrt{3}}{1}= 4\sqrt{3}.

Let's put everything back into the first mentioned identity.

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}

\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}

\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}

4 0
3 years ago
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