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Troyanec [42]
3 years ago
10

Phill weighs 120 pounds and is gaining ten pounds each month. Phil weighs 150 pounds and is gaining 4 pounds each month. How man

y months, m, will it take for Bill to weigh the same as Phil?
Mathematics
1 answer:
nikdorinn [45]3 years ago
4 0

Answer:

5 months

Step-by-step explanation:

You had a typo and said that Phil weighed both amounts so I'm assuming that the second amount is Bill's weight

We can use LCM for this

So lets make a list for both of them

Phil:120, 130, 140, 150, 160, 170, 180, 190, 200

Bill: 150, 154, 158, 162, 166, 170, 174

As you can see here, after 5 months, they will have the same weight.

We can check our work to.

lets do 120+(10*5)=170

and 150+(4*5)=170

I hope this makes sense

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Elza [17]

Answer: only (x-y=4) and (x+y=4).


Step-by-step explanation:

Notice target point has y=0, so all terms with y are zero. Then all 6 equations reduce to m x = k, for various m and k. So calculate 4×m and compare to k six times.


x - y = 4 4=4 yes

-x - y = 4 -4=4 no

2x - y = 7 8=7 no

x + y = 4 4=4 yes

2x + y = 7 8=7 no

2x + y = -7 8=-7 no.

6 0
3 years ago
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Translate into an equation or inequality to solve: Two more than twice the sum of a number and 12 is greater than the difference
tamaranim1 [39]
Let say the number is 0.5
Then, sum of this number will be 0.5+0.5=1
Now two is twice of 1.
Hence, it proved to be correct.
And difference between 5 and 1 is 4 which is less than 12. It also match the question condition.

Answer: The number is 1.
3 0
3 years ago
Which functions have an additive rate of change of 3? Select TWO options
Pachacha [2.7K]

Answer:

Second table.

Step-by-step explanation:

A function has an additive rate of change if there is a constant difference between any two consecutive input and output values.

The additive rate of change is determined using the slope formula,

m =  \frac{y_2-y_1}{x_2-x_1}

From the first table we can observe a constant difference of -6 among the y-values and a constant difference of 2 among the x-values.

m =  \frac{ - 9- - 3}{4-2}  =  - 3

For the second table there is a constant difference of 3 among the y-values and a constant difference of 1 among the x-values.

The additive rate of change of this table is

m =  \frac{ - 1 -  - 4}{3 - 1}  = 3

Therefore the second table has an additive rate of change of 3.

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its C

Step-by-step explanation:

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Bond [772]

The solution to these equations is either x < 0 or x > 3.

In order to find them, we need to solve each equation separately. Let's start with the first one.

2x - 1 < -1 -----> Add 1 to both sides

2x < 0 -----> Now divide each side by 2.

x < 0

Now let's look at the second one.

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x > 3 (Notice that the sign changes direction because we divided by a negative)

When you have an "or" statement, you'll wind up with two answers, so we use both of these.

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