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ivann1987 [24]
3 years ago
5

Select THREE equations whose graphs are straight lines

Mathematics
1 answer:
NARA [144]3 years ago
3 0

Answer:

A, C and D

Step-by-step explanation:

A:  y = 7 is a straight, horizontal line

B: is a parabola.  Discard

C:  (cannot read denominator)  This is a straight line thru the origin.

D:  This is a straight line.

E:  This is a parabola, due to to the x^2 term.  Discard

F:  This is a circle.  Discard

Straight lines here are A, C and D

You might be interested in
For t=2, 15t - 2t is equal to
weqwewe [10]

Answer:

26

Step-by-step explanation:

=15(2) - 2(2)

=30 - 4

=26

6 0
3 years ago
Write an equation in slope-intercept form of the line that passes through (7, 2) and (2, 12)
Anika [276]

Step-by-step explanation:

the general slope intercept form is

y = ax + b

a is the slope, b is the y-intercept (when x=0).

the slope is the ratio y/x indicating how much y changes, when x changes for a certain amount of units when going from one point to another.

going from (7, 2) to (2, 12) :

x changes by -5 (from 7 to 2), y changes by +10 (from 2 to 12).

so, the slope is 10/-5 = -2

therefore, we have already

y = -2x + b

we get b by using the coordinates of one of the points as x and y. e.g. (7, 2)

2 = -2×7 + b = -14 + b

16 = b

so, the full line equation is

y = -2x + 16

4 0
3 years ago
The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in gr
raketka [301]

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass <em>M</em>₀, the mass after 12 hours is half that:

1/2 <em>M</em>₀ = <em>M</em>₀ exp(12<em>k</em>)

for some decay constant <em>k</em>. Solve for this <em>k</em> :

1/2 = exp(12<em>k</em>)

ln(1/2) = 12<em>k</em>

<em>k</em> = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass <em>M</em>₀, the mass <em>M</em> remaining after time <em>t</em> is given by

<em>M</em> = <em>M</em>₀ exp(<em>kt </em>)

So if <em>M</em>₀ = 590 g and <em>t</em> = 36 h, plugging these into the equation with the previously determined value of <em>k</em> gives

<em>M</em> = 590 exp(36<em>k</em>) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74

6 0
3 years ago
Use set builder notation to write each of the following sets 45,55,65,75
butalik [34]

Answer: \{ x: x =5(2n+1) ,4\leq n\leq 7 ,x,n\in N \}

Explanation:

As we know that set is the collection of elements and set builder is the method of defining the set with the help of specific property in which we can use arithmetic operations as well

Generally,

It is the general form of any set in which we can construct the complete set  with the help of equation in the given in set builder form.

Since we have the elements : 45,55,65,75

So, the set builder form will be { x: x =5(2n+1) ,4≤n≤7,x,n∈ N }

We will read it as it is the value of x which will be find with the help of given equation i.e. x=5(2n+1) where n goes from 4 to 7 only and x and n are natural numbers only.

3 0
3 years ago
Can someone help me?
Alika [10]

Answer:

I think it might be 6

Step-by-step explanation:

You should probably wait for another answer, I did this in Math last year so I think it's 6 but I can't really remember sorry ):

4 0
3 years ago
Read 2 more answers
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