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Alex787 [66]
3 years ago
6

There are 3 consecutive odd integers that sum to 45. What is the least of the 3 integers?

Mathematics
1 answer:
Anna71 [15]3 years ago
5 0

Step-by-step explanation:

Let n = a number and 2n = an even number, then ...

Let 2n + 1 = the first of three consecutive odd numbers.

Let 2n + 3 = the second of three consecutive numbers.

Let 2n + 5 = the third of three consecutive odd numbers.

Since it's given that the sum of our three consecutive odd numbers is 45, then we can write the following equation to be solved for the variable n:

(2n + 1) + (2n + 3) + (2n + 5) = 45

2n + 1 + 2n + 3 + 2n + 5 = 45

Collecting like-terms on the left, we get:

6n + 9 = 45

Now, subtracting 9 from both sides in order to begin isolating the unknown, n, on the left side:  

6n + 9 - 9 = 45 - 9

6n + 0 = 36

6n = 36

Now, divide both sides by 6 in order to solve the equation for n:

(6n)/6 = 36/6

(6/6)n = 36/6

(1)n = 6

n = 6

Therefore, ...

2n + 1 = 2(6) + 1

= 12 + 1

= 13

2n + 3 = 2(6) + 3

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18 tens 20 ones = ______hundreds
Elis [28]

Answer:

200

Step-by-step explanation:

18 × 10 = 180

20 × 1 = 20

180 + 20 = 200

3 0
4 years ago
The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?
11111nata11111 [884]

we have

A(-2, 2),B(6, 2),C(0, 8)

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

P=AB+BC+AC

and

the area of the triangle is equal to

A=\frac{1}{2}*base *heigth

in this problem

base=AB\\heigth=DC

we know that

The distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Step 1

<u>Find the distance AB</u>

A(-2, 2),B(6, 2)

Substitute the values in the formula

d=\sqrt{(2-2)^{2}+(6+2)^{2}}

d=\sqrt{(0)^{2}+(8)^{2}}

dAB=8\ units

Step 2

<u>Find the distance BC</u>

B(6, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-6)^{2}}

d=\sqrt{(6)^{2}+(-6)^{2}}

dBC=6\sqrt{2}\ units

Step 3

<u>Find the distance AC</u>

A(-2, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0+2)^{2}}

d=\sqrt{(6)^{2}+(2)^{2}}

dAC=2\sqrt{10}\ units

Step 4

<u>Find the distance DC</u>

D(0, 2),C(0, 8)

Substitute the values in the formula

d=\sqrt{(8-2)^{2}+(0-0)^{2}}

d=\sqrt{(6)^{2}+(0)^{2}}

dDC=6\ units

Step 5

<u>Find the perimeter of the triangle</u>

P=AB+BC+AC

substitute the values

P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units

P=22.81\ units

therefore

The perimeter of the triangle is equal to 22.81\ units

Step 6

<u>Find the area of the triangle</u>

A=\frac{1}{2}*base *heigth

in this problem

base=AB=8\ units\\heigth=DC=6\ units

substitute the values

A=\frac{1}{2}*8*6

A=24\ units^{2}

therefore

the area of the triangle is 24\ units^{2}

4 0
3 years ago
Read 2 more answers
0I NEED HELP PLS........
emmasim [6.3K]

Answer:

Not equivalent

Step-by-step explanation:

Distribute the negative in the parentheses. It basically multiplying everything inside by -1.

( \frac{1}{2}  + 6x + 3)

add like terms

6x + 3 \frac{1}{2}

Not equivalent since the 6x is positive

6 0
3 years ago
Read 2 more answers
Diego is solving the equation x^2-12x = 21
uysha [10]

Answer:

The solutions to the quadratic equations will be:

x=\sqrt{57}+6,\:x=-\sqrt{57}+6

Step-by-step explanation:

Given the expression

x^2-12x\:=\:21

Let us solve the equation by completing the square

x^2-12x\:=\:21

Add (-6)² to both sides

x^2-12x+\left(-6\right)^2=21+\left(-6\right)^2

simplify

x^2-12x+\left(-6\right)^2=57

Apply perfect square formula: (a-b)² = a²-2ab+b²

i.e.

x^2-12x+\left(-6\right)^2=\left(x-6\right)^2

so the expression becomes

\left(x-6\right)^2=57

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solve

x-6=\sqrt{57}

add 6 to both sides

x-6+6=\sqrt{57}+6

Simplify

x=\sqrt{57}+6

also solving

x-6=-\sqrt{57}

add 6 to both sides

x-6+6=-\sqrt{57}+6

Simplify

x=-\sqrt{57}+6

Therefore, the solutions to the quadratic equation will be:

x=\sqrt{57}+6,\:x=-\sqrt{57}+6

5 0
3 years ago
What is an algebraic expression for the following word phrase: "6 times the difference of g and 3"
myrzilka [38]
6xG-3
Thats the answer
8 0
4 years ago
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