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harkovskaia [24]
3 years ago
5

The sum of three consecutive consecutive integers is 141 what is the smallest integer

Mathematics
2 answers:
Romashka [77]3 years ago
5 0
Let x=the smallest integer, x+1=the second smallest, x+2=the greatest.
x+x+1+x+2=141
3x+3=141
3x=138
x=46
46 is the smallest integer
zvonat [6]3 years ago
5 0

Our three consecutive integers can be represented as

X → first integer

X + 1 → second integer

X + 2 → third integer

If the sum of our three consecutive integers is 141, our equation will read...

X + X + 1 + X + 2 = 141

Now, we can simplify on the left. Once we simplify, our equation will now read

3x + 3 = 141

      -3     -3  ← subtract 3 on both sides

      3x = 138 ← divide by 3 on both sides

       <u>X = 46</u> ← smallest integer

Therefore, the smallest integer is 46.

       

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3 years ago
Find the distance between the two points.<br> (-6, – 7), (0,0)
Elden [556K]

Answer:

√85

Step-by-step explanation:

for this, you will probably need to draw a graph. Plot the points, and then you will notice that you can draw a triangle with one side being the x-axis ( go on to (-6,0) then drop down unitll you hit the point (-6;-7)

// use the pythagoream theorem to find the length

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the distance of the leg that is dropping down is -7 (calculate 0-(-7))

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8 0
3 years ago
PLEASE HELP ASAP! I don’t recall how to do this!
MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

(1.05)^x=1.885

In order to solve for x, we have to get it out from its position of an exponent.  Do that by taking the natural log of both sides:

ln(1.05)^x=ln(1.885)

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

x=\frac{ln(1.885)}{ln(1.05)}

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

ln(x*x^2)=5

Simplifying gives you

ln(x^3)=5

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

ln(x)=\frac{5}{3}

Take the inverse ln by raising each side to e:

e^{ln(x)}=e^{\frac{5}{3}}

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

\frac{1}{2}=e^{.1x}

"Undo" that e by taking the ln of both sides:

ln(.5)=ln(e^{.1x})

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

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x = -6.931471806

Question d. is a bit more complicated than the others.  Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

(2^2)^x-6(2)^x=-8

Now we will bring over the -8 by adding:

(2^2)^x-6(2)^x+8=0

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution.  Let's let u=2^x

When we do that, we can rewrite the polynomial as

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This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor.  Now we have to put it back in:

2^x=4,2^x=2

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

2^2=2^x

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

2^x=2^1

Now that the bases are the same, we can say that

x = 1

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Stolb23 [73]
I also got the answer of 144 too
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