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Contact [7]
3 years ago
15

Plz help me I’m stuck on this question

Mathematics
2 answers:
Ksju [112]3 years ago
8 0
The equation is N= 172-42÷2.
igomit [66]3 years ago
4 0
172/2-45=n is your answer
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Pls help me fast and give me the steps pls<br> 1+7p-40=80
valentinak56 [21]

Answer:

17

Step-by-step explanation:

1+7p-40=80

-39+7p=80 Move constant to the right

7p=80+39 add the numbers

7p=119 divide both sides by 7

4 0
3 years ago
Read 2 more answers
This is the table of values of a function. Write the output when the input is n.
denis23 [38]

Answer:

n-2

Step-by-step explanation:

1. Find what has been done to the input value to give the output value. Here, we can see that each input value is two more than each output value, so two has been subtracted from each input value to give the output. This means that the rule for this function is: <em>take the input value and subtract two.</em>

2. Express this in terms of n. We can then express this in terms of n (using n to show the rule that we have found). Since n is the input value we have, to find the output when the input is n, we subtract two from n. This gives us our answer of n-2.

4 0
4 years ago
Help please 30 points
tino4ka555 [31]

Answer:

(22,0)

Step-by-step explanation:

y=mx+b

y=-2/3x+44/3

0=-2x/3+44/3

x=22

(22,0)

5 0
3 years ago
Convert x - 5 = 0 to a polar equation
dolphi86 [110]

Answer:

(

5

,

π

)

Step-by-step explanation:

−

5

=

r

cos

θ

88

[

1

]

0

=

r

sin

θ

8888

[

2

]

Squaring [1] and [2]:

25

=

r

2

cos

2

θ

0

=

r

2

sin

2

θ

Adding [1] and [2]:

25

=

r

2

cos

2

θ

+

r

2

sin

2

θ

Factor:

25

=

r

2

(

cos

2

θ

+

sin

2

θ

)

25

=

r

2

⇒

r

=

±

5

( use

r

=

5

)

4 0
4 years ago
A set of elementary school student heights are normally distributed with a mean of 105105105 centimeters and a standard deviatio
steposvetlana [31]

Answer:

The proportion of student heights that are between 94.5 and 115.5 is 86.64%

Step-by-step explanation:

We have a mean \mu = 105 and a standard deviation \sigma = 7. For a value x we compute the z-score as (x-\mu)/\sigma, so, for x = 94.5 the z-score is (94.5-105)/7 = -1.5, and for x = 115.5 the z-score is (115.5-105)/7 = 1.5. We are looking for P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664. Therefore, the proportion of student heights that are between 94.5 and 115.5 is 86.64%

4 0
4 years ago
Read 2 more answers
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