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Iteru [2.4K]
2 years ago
5

Find the x-intercept and the y-intercept of 2x + 6y =24

Mathematics
1 answer:
viktelen [127]2 years ago
4 0

Answer:

x-intercept = 12, y-intercept = 4

Step-by-step explanation:

To find the x and y intercept, substitute 0 in for both values.

<u>Substitute 0 into x:</u>

2(0) + 6y = 24

6y = 24

<u>Divide each side by 6:</u>

y = 4

<u>Substitute 0 into y:</u>

2x + 6(0) = 24

2x = 24

<u>Divide each side by 2:</u>

x = 12

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Simplify The Expression 16(1/2x)^4
Bond [772]
16(1/2x)^4
Answer: X^4
8 0
2 years ago
I dont know what I'm doing! could you help me​
photoshop1234 [79]

Answer:

c = 24.34

Step-by-step explanation:

Here, we can use the cosine rule

Generally, we have this as:

a^2 = b^2 + c^2 - 2bcCos A

12^2 = 14^2 + c^2 - 2(14)Cos 19

144 = 196 + c^2 - 26.5c

c^2 - 26.5c + 196-144 = 0

c^2 - 26.5c + 52 = 0

We can use the quadratic formula here

and that is;

{-(-26.5) ± √(-26.5)^2 -4(1)(52)}/2

(26.5 + 22.23)/2 or (26.5 - 22.23)/2

24.37 or 2.135

By approximation c = 24.34 will be correct

8 0
3 years ago
If 95 % of the students are present in a school and the number of absent students is 25, find the total number of students in th
Karolina [17]
100 minus 95 equals 5
25 equals 5%
100 divided by 5 equals 20
20 times 25 equals 500
500 minus 25 equals 475
Answer:475 students present/ 500 students total in school
8 0
3 years ago
Please answer this for me
Anvisha [2.4K]

Answer:

70

Step-by-step explanation:

you add 95 and 15 and get 110 then you do 180-110 getting 70

6 0
3 years ago
The air force reports that the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a
Anon25 [30]

Answer:

Step-by-step explanation:

<em>Given that X -  the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a standard deviation of 2.7 inches.</em>

<em />

<em>Height of male pilot = 74.2 inches</em>

<em />

<em>We have to find the percentile</em>

<em />

<em>X = 74.2</em>

<em />

<em>Corresponding Z score = 74.2-72.6 = 1.6</em>

<em />

<em>P(X<174.2) = P(Z<1.6) = 0.5-0.4452=0.0548=5.48%</em>

<em />

<em>i.e. only 5% are below him in height.</em>

<em />

<em>Thus the malepilot is in 5th percentile.</em>

4 0
2 years ago
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