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Marina86 [1]
3 years ago
9

For the following exercises, determine whether the relation is a function

Mathematics
1 answer:
siniylev [52]3 years ago
3 0
Ummmmmmmmmmmmmmmmmmmmmmm
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If h(z)=13z2–30z+27, use synthetic division to find h(3).
Blababa [14]

Answer:

h=13z-30+ 27/z

Step-by-step explanation:

4 0
3 years ago
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Which ordered pairs are in the solution set of the system of linear inequalities?
Digiron [165]

Answer:

c.  (5,-2), (3, 1), (4,2).

Step-by-step explanation:

The solutions are in the dark red area, with some on the continuous blue line but none on the red dotted line.

3 0
4 years ago
Sara is ordering t-shirts for her school. Sally's Shirt Shop charges $8.50 for each shirt and a one-time set up fee of $60. Shau
tensa zangetsu [6.8K]

Answer:

A ) 8.50x + 11 < 24x + 60

B ) 8.50x + 24 < 11x + 60

C ) 11x + 24 < 8.50x + 60

D ) 8.50x + 60 < 11x + 24

Step-by-step explanation:

8 0
3 years ago
Please answer the question in the picture and show your work! I got tan= square root of 2 but I’m not sure if that’s right and i
melomori [17]

Answer:

tan(θ) is undefined.

Step-by-step explanation:

Recall that tan(θ) = sin(θ) / cos(θ). We are given that sin(θ) = -1. Hence:

\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}=-\frac{1}{\cos\theta}

Since 1 / cos(θ) = sec(θ):

\tan\theta=-\sec\theta

We can square both sides:

\tan^2\theta=\sec^2\theta

From the Pythagorean Identity:

\tan^2\theta+1=\sec^2\theta

Substitute:

\tan^2\theta+1=\tan^2\theta

So:

1\neq0

Since we acquire an untrue statement, no solutions exist.

If needed, refer to the unit circle. Recall that sin(θ) only equals -1 when θ = 3π/2 (on the interval [0, 2π)). At 3π/2, cos(θ) = 0. Since tan(θ) = sin(θ) / cos(θ), tan(θ) is undefined whenever sin(θ) = -1 (or 1).

6 0
3 years ago
Given the vertices of ∆ABC are A (2,5), B (4,6) and C (3,1), find the vertices following each of the transformations FROM THE OR
djyliett [7]
The answer

a) the main rule for transformations <span>Rx-axis is as following
 original (x, y) --------------- the final coordinates become  (x, -y)

so the answer is  </span><span>A' (2, -5), B' (4,-6) and C' (3,-1)

b) just put the figure of y= 3, and check the value after projection

it is </span> A' (2, 1), B' (4,0) and C' (3,4)

c) <span>T<-2,5>, the main rule is as following

T(a, b) ----- applying on (x, y) is  (a+x, b+y) for the transformed points

in our case, </span><span>A' (0,10), B' (2,11) and C' (1,6)

d) the same method as above for </span><span>d. T<3,-6>, we found
</span><span>A '(5,-1), B '(7,0) and C ' (6,-5)


</span>
3 0
3 years ago
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