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icang [17]
2 years ago
11

Graph the line with slope 3 and intercept 1 X​

Mathematics
1 answer:
mr_godi [17]2 years ago
7 0
Plot a point at (0,-1)
Then from that point, go down 3 points, and to the right one point. Repeat that until you hit the bottom of the graph or run out of room. From the first point (0,-1), go up 3 and to the left 1 until you run out of room on the graph. Should be a straight line
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If sec(x)=25/24, find cotangent (x)
nadezda [96]
The answer is 1.042 because you have to divide 25 by 24
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Use substitution to solve the system of equations.<br> x-2y=1<br> 3x-6y=3
Ivahew [28]
Change first equation
x - 2y = 1
add 2y to both side
x = 2y + 1
Plug into second equation
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7 0
3 years ago
Write the mixed number 5 2/25 as a percent.
Nataliya [291]

Answer:

I'm assuming it's like this : 5*2/25 which will be 0.4, which is 40%

3 0
3 years ago
For the graph, what is a reasonable constraint so that the function is at least 300?
blsea [12.9K]

Answer:

C) 0 ≤ x ≤ 25

Step-by-step explanation:

We are supposed to find a reasonable constraint so that the function is at least 300 i.e. the value of x at which f(x) is greater or equal to 300

A)x ≥ 0

Refer the graph

At x = 0

f(x)=300

On increasing the value of x , f(x) increases but at x = 12 it starts decreasing

So, x ≥ 0 can also have f(x)<300

So, Option A is wrong

B)−5 ≤ x ≤ 30

At x = -5

f(x) = 100

So, Option B is wrong since we require f(x) is greater or equal to 300

c)0 ≤ x ≤ 25

At x = 0

f(x)=300

At x = 12 , it starts decreasing

At x = 25

f(x)=300

So, The value of f(x) is at least 300 when 0 ≤ x ≤ 25

D)All real numbers

At x = 30

f(x)=0

But we require f(x) greater or equal to 300

Hence Option C is true

8 0
3 years ago
Given the following triangle, if a = 2 and b = 3, find c.<br> √5<br> √10<br> √13
Tems11 [23]
We can use pythagorean theorem:

a^2 + b^2 = c^2

4 + 9 = c^2

13 = c^2

sqrt of 13 = c

Hope this helps!
4 0
3 years ago
Read 2 more answers
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