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Oduvanchick [21]
3 years ago
14

Can some one help me, its either positive,negative,zero, or undefined

Mathematics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

So the slope is a negative slope, the equation to go with the graph would be...

y = 1x   or   y = 1x + 0

I hope this helps

Step-by-step explanation:

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A cafeteria manager needs to know how many apples and bananas to to order. He asked students to choose either an apple or a bana
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Step-by-step explanation:

3/4 of total = 45

therefore, 1/4 took bananas

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so we get 45/3 = 15 people picking bananas

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3 years ago
Answer my questions IM FAILING MATH AND NO OKES HELPING ME
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Answer:

Hi!

Around 2208.97 dollars

Step-by-step explanation:

A=P(1+r/n)^nt This is the compound interest rate equation.

According to the problem, Principal is 2000 with annual rate of 2.5%, it was being compounded semiannually(twice a year).

Thus,

P=2000.  r/n=2.5%/2 nt=2*4, 8

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3 years ago
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What is the distance from the pair of points (7,3) and (-1,-4)
mafiozo [28]

Answer:

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4 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!

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<> because it means greatest and least only
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