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Schach [20]
3 years ago
6

Can someone help me please

Mathematics
1 answer:
Misha Larkins [42]3 years ago
4 0

Answer:

-28

Step-by-step explanation:

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HELPPPPPP ASAP I REALLY NEED HELP 20points
Nezavi [6.7K]

Answer:

x^2y^{11} / 16

Step-by-step explanation:

Hope this helped!

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Don has an album that holds 900 coins . Each page of the album holds 9 coins . If 78​% of the album is​ empty, how many pages ar
laila [671]

Answer:  22% of the pages are filled up

Step-by-step explanation: 900/9= 100 78% of 100 is 78

22+78=100 therefore reaching 100 percent. since 78% is empty that means 22% is being used.

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3 years ago
What is 7 and a half hours + 6 and 3/4 hours + 7 and 1/3 hours ?
Korolek [52]
7 and 1/2+6 and 3/4+7 and 1/3=7+1/2+6+3/4+7+1/3=7+6+7+1/2+3/4+1/3=20+1/2+3/4+1/3
convert bottom nubmer to same (12)
1/2 times 6/6=6/12
3/4 times 3/3=9/12
1/3 times 4/4=4/12
20+6/12+9/12+4/12=20+19/12=20+1+7/12=21+7/12=21 and 7/12
answer is 21 and 7/12 hour or 21.58 hour or 21 hour and 35 minutes
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3 years ago
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Please answer correctly !!!!!!!!! Will mark Brianliest !!!!!!!!!!!!
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Line NM is 9 and the measure of angle MLN is 58 degrees. This is because both triangles are equal in length, size, and angle measures. The triangle to the right is just rotated and reflected.
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3 years ago
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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
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