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2 years ago

15

The sum of three consecutive odd numbers is 147. what is the smallest of the three numbers?

1 answer:

Andrew [12]2 years ago

3 0

Let x be the smallest odd number.
x+(x+2)+(x+4)=147
3x+6=147
x=47
therefore the smallest odd no. is 47.

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Match the following items by evaluating the expression for x = -2.

Sloan [31]

Answer:

1) $-4$

2) $-3$

3) $1$

4) $-2$

5) $4$

Step-by-step explanation:

According to the information given in the exercise, the value of "x" is:

$x = -2$

Then:

1) You must substitute $x = -2$ into the expresion $x-2$ :

$=(-2)-2$

Now, you must evaluate:

$=-4$

2) You have to substitute the given value of "x" into the second expression:

$x-1=(-2)-1$

Finally, evaluating, you get the following result:

$=-3$

3) According to the Zero exponent Rule, any non-zero number with an exponent of zero is equal to 1.

Then, substituting $x = -2$ into the expression $x^0$ and then evaluating, you get:

$x^0=(-2)^0=1$

4) By definition:

$a^1=a$

Then, substituting $x = -2$ into $x^1$ and evaluating, you get:

$x^1=(-2)^1=-2$

5) Since:

$a^2=a*a$

When you substitute the given value of "x" into the last expression and then you evaluate, you get:

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4 0

2 years ago

Please help! Give the answer and how you got it

Ronch [10]

The distance between the points C(-3,1) and D(5,6) is 10.3 units.

<u>Step-by-step explanation:</u>

Let us consider CD is a line segment.

The point C is (-3,1) and D is (5,6).

The Cartesian formula for line is used to find the length between the CD.

Distance between two points = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

The points are (-3,1) and (5,6).

= $\sqrt{(6-(-3))^2+(6-1)^2}$ .

= $\sqrt{(6+3)^2+(6-1)^2}$

= $\sqrt{9^2+5^2}$ .

= $\sqrt{81+25}$ .

= $\sqrt{106}$ .

=10.295 ≈ 10.3.

Distance between two points 10.3 units.

7 0

3 years ago