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konstantin123 [22]
3 years ago
7

Favorite vine? giving brainliest <3

Mathematics
2 answers:
Naya [18.7K]3 years ago
5 0

Answer:lOoK aT aLl ThEsE cHiCkEnS

Step-by-step explanation:

svlad2 [7]3 years ago
5 0

Answer:

Women singing to songs

Step-by-step explanation:

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A rectangle has width (2x + 3) and length (2x - 4). What is the AREA of the rectangle?
Ksivusya [100]

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85

Step-by-step explanation:

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Does is still show up black? Hope fully you can see it now, but please help :’)
dangina [55]
No we can see it now
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What is 1/7^9 × 1/7^-6
artcher [175]
\bf \cfrac{1}{7^9}\cdot \cfrac{1}{7^{-6}}\implies \cfrac{1\cdot 1}{7^9\cdot 7^{-6}}\implies \cfrac{1}{7^{9-6}}\implies \cfrac{1}{7^3}\implies \cfrac{1}{343}
5 0
3 years ago
20 POINTS!! ASAP, PLS SHOW WORK TYY
Sergeeva-Olga [200]

Answer:

\sin(\theta)=-\sqrt5/5\text{ and } \csc(\theta)=-\sqrt5\\\cos(\theta)=2\sqrt5/5\text{ and } \sec(\theta)=\sqrt5/2\\\tan(\theta)=-1/2\text{ and } \cot(\theta)=-2

Step-by-step explanation:

First, let's determine which quadrant our angle θ lies in.

Remember ASTC, where:

Everything is positive in QI,

Only sine (and cosecant) is positive in QII,

Only tangent (and cotangent) is positive in QIII,

And only cosine (and secant) is positive in QIV.

Since our tangent is negative, and our cosine is positive, this means that our θ <em>must</em> be in QIV.

In QIV, sine is negative, tangent is negative, and cosine is positive.

With that, let's figure out the remaining trig ratios.

We know that:

\tan(\theta)=-1/2

Remember that tangent is the ratio of the opposite side to the adjacent side.

Let's figure out our hypotenuse using the Pythagorean Theorem:

a^2+b^2=c^2

Substitute 1 for a and 2 for b (we can ignore the negative since we're squaring anyways). This yields:

(1)^2+(2)^2=c^2

Square:

1+4=c^2

Add:

c^2=5

Take the square root:

c=\sqrt{5}

So, our square root is √5.

So, our three sides are: Opposite=1, Adjacent=2, and Hypotenuse=√5.

Sine and Cosecant:

Remember that:

\sin(\theta)=opp/hyp

Substitute 1 for the opposite and √5 for the hypotenuse. This yields:

\sin(\theta)=1/\sqrt5

Rationalize:

\sin(\theta)=\sqrt5/5

And since our angle is in QIV, we add a negative:

\sin(\theta)=-\sqrt5/5

Cosecant is simply the reciprocal of sine. So:

\csc(\theta)=-\sqrt5

Cosine and Secant:

Remember that:

\cos(\theta)=adj/hyp

Substitute 2 for the adjacent and √5 for the hypotenuse. This yields:

\cos(\theta)=2/\sqrt5

Rationalize:

\cos(\theta)=2\sqrt5/5

Since our angle is in QIV, cosine stays positive.

Secant is the reciprocal of cosine. So:

\sec(\theta)=\sqrt5/2

Tangent and Cotangent:

We were given that:

\tan(\theta)=-1/2

To find cotangent, flip:

\cot(\theta)=-2

And we're done!

7 0
3 years ago
I need help with this question
jeka57 [31]

Answer:

Step-by-step explanation:

EF = DC, THAT'S WRONG

so everything correct but C

and done! hope you learn something new!

7 0
3 years ago
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