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andriy [413]
2 years ago
10

Guided Practice

Mathematics
2 answers:
nlexa [21]2 years ago
6 0

Answer:

c. (0,0)

Step-by-step explanation:

The midpoint between 9 and -9 is 0, because 9 to 0 is 9, and -9 to 0 is 9.

The midpoint between 12 and -12 is 0 as well, because 12 to 0 is 12, -12 to 0 is also 12.

KATRIN_1 [288]2 years ago
5 0

Answer:

i would think its C

Step-by-step explanation:

the other options dont make sense lol

hope this is correct and helps :)

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Jane gets paid $120 for 8 hours. How much does she get paid an hour?​
AleksAgata [21]

Answer:

$15 an hour

Step-by-step explanation:

To find this, we divide 120 by 8.

120 ÷ 8 = 15

So Jane gets $15 an hour

I hope this helped, please mark Brainliest, thank you!

8 0
2 years ago
In a right triangle, <A and <B are acute. Find the value of sin A, when tan A = 8/15​
finlep [7]

Answer:sinA=8/17

Step-by-step explanation:

Using Pythagoras principle

x=√(15^2+8^2)

x=√(15x15+8x8)

x=√(225+64)

x=√(289)

x=17

SinA=opposite/hypothenus

SinA=8/17

3 0
3 years ago
What value of c makes x^2+6+c a perfect square trinomial ?
Citrus2011 [14]

According to this, then:

6x = 2*sqrt(c)*x

6 = 2*sqrt(c)

3 = sqrt(c)

c = 3^2 = 9

Hope this helps!

6 0
3 years ago
Read 2 more answers
Provide the reflected ordered pair for each of the given ordered pairs. Reflect the ordered pair ( − 4 , 5 ) (-4, 5) across the
levacccp [35]

Answer:

a) (4,5)

b) (0,-3)

Step-by-step explanation:

We have to perform the following reflection over given ordered pair.

a) Reflect the ordered pair (-4,5) across the y-axis

Reflection over y-axis:

r: (x,y)\rightarrow (-x,y)

Thus, (-4,5) will be reflected over y-axis as

r: (-4,5)\rightarrow (-(-4),5) = (4.5)

b) Reflect the ordered pair (0,3) across the y-axis

Reflection over x-axis:

r: (x,y)\rightarrow (x,-y)

Thus, (0,3) will be reflected over x-axis as

r: (0,3)\rightarrow (0,-(3)) = (0,-3)

6 0
3 years ago
Use the Ratio Test to determine whether the series is convergent or divergent.
natka813 [3]

Answer:

The series is absolutely convergent.

Step-by-step explanation:

By ratio test, we find the limit as n approaches infinity of

|[a_(n+1)]/a_n|

a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)

a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)

[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]

= |-3n³/2(n+1)³|

= 3n³/2(n+1)³

= (3/2)[1/(1 + 1/n)³]

Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity

= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity

= 3/2 × 1

= 3/2

The series is therefore, absolutely convergent, and the limit is 3/2

3 0
2 years ago
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