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svp [43]
3 years ago
5

Whst is the answer to 1 3/4 - 7/10 = ?

Mathematics
2 answers:
alukav5142 [94]3 years ago
5 0
Turn 1 3/4 into a decimal which equal 1.75. Because 10/10 = 1.00
Same thing with 7/10 = 0.70

Take 1.75 - 0.70 = 1.05
Verdich [7]3 years ago
5 0
Change them into improper fraction with a common  denominator is easier
So you should get 70/40-28/40
The answer is 42/40 or simplified is 1  1/20
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En que consiste la jerarquía de operaciones?
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Answer:

La jerarquía de operaciones es un método para resolver operaciones con múltiples operadores; saber realizarla te servirá para resolver los diversos problemas que te presenten en tu examen CENEVAL EXANI-II.  En primer lugar, se deben resolver las operaciones de potencias y raíces.

Step-by-step explanation:

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HELP ASAP. Worth 10 points
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2 years ago
If point E exists such that its x-coordinate is 12 and ED is parallel to AC, then what is the y-coordinate of point E? Show how
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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!

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