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Lelu [443]
3 years ago
12

DOUBLE POINTS for answering the two questions

Mathematics
1 answer:
Margarita [4]3 years ago
6 0

Answer:

i have the same question

Step-by-step explanation:

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Which graph represents the solution set of the system of inequalities?
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The answer to this is D
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A.0 B.-7 C.1. D.undefined Whst one is is pls help
Musya8 [376]

Answer:

undefined

Step-by-step explanation:

The line on the graph is completely vertical, which is characteristic of an undefined graph.

4 0
3 years ago
If fence poists are to be placed in a row 7 feet apart, how many posts are needed for 210 feet of fence?
ollegr [7]

Answer:

31

Step-by-step explanation:

210 ft ÷ 7 ft = 30

There are 30 7-ft lengths of fence.

Think of putting a post at the end of each 7-ft length. That means you need 30 posts. Now you must add 1 post for the beginning.

Answer: 31

8 0
2 years ago
Plz help due tomorrow
sp2606 [1]

Answer:

21y^2+11.2

Step-by-step explanation:

4-3y(-7y)+7.2

combine like terms

-3y (-7y) = 21y^2

add 4 to 7.2

11.2

21y^2 + 11.2

8 0
2 years ago
WILL GIVE BRAINIEST IF DONE IN 10 MINUTES!!!!!
Mama L [17]

Answer:

\sin(\theta)=-12/13\text{ and } \csc(\theta)=-13/12\\\cos(\theta)=-5/13\text{ and } \sec(\theta)=-13/5\\\tan(\theta)=12/5\text{ and } \cot(\theta)=5/12

Step-by-step explanation:

So we know that:

\sin(\theta)=-12/13\text{ and } \sec(\theta)

This tells us that our angle θ is in Quadrant III. This is because, recall ASTC. In QI, everything is positive. This is not the case here since sine and secant are both negative.

In QII, <em>only</em> sine is positive. Since sine is negative, θ can't be in QII.

In QIII, <em>only</em> tangent (and cotangent) is positive. All other are negative. So, θ is in QIII.

And just to check, in QIV, <em>only</em> cosine (and secant) is positive. Since we are told that secant is less than 0 (in other words, negative), θ can't be in QIV.

Now that we know that θ is in QIII, we know that sine and cosecant is negative; cosine and secant is negative; and tangent and cotangent is positive.

Now, let's find the ratios. Remember what sine tells us. Sine gives us the ratio of the <em>opposite</em> side to the <em>hypotenuse</em>. With this information, let's find the adjacent side:

So, the opposite is 12 (ignore the negative for now) and the hypotenuse is 13. Therefore, we can use the Pythagorean Theorem:

a^2+b^2=c^2

Substitute 12 for a and 13 for c:

12^2+b^2=c^2

Now, solve for b to find the adjacent side. Square the values:

144+b^2=169

Subtract 144 from both sides:

b^2=25

Square root:

b=\pm 5

We can ignore the negative and just take the positive. While the answer is technically -5, we don't need to do that. So, our adjacent side is 5.

Therefore, with respect to angle θ, our opposite is 12, our adjacent is 5, and our hypotenuse is 13. With this, we can compute the other trig ratios. Let's start with the 3 main ones:

Sine:

\sin(\theta)

As given, this is -12/13. So:

\sin(\theta)=-12/13

Cosine:

\cos(\theta)=adj/hyp

Substitute 5 for the adjacent and 13 for the hypotenuse. So:

\cos(\theta)=5/13

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

\cos(\theta)=-5/13

Tangent:

\tan(\theta)=opp/adj

Substitute 12 for opposite and 5 for adjacent. So:

\tan(\theta)=12/5

And since our angle is in QIII, tangent stays positive.

To find the other three, simply flip the previous fractions:

\csc(\theta)=-13/12\\\sec(\theta)=-13/5\\\cot(\theta)=5/12

5 0
3 years ago
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