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fenix001 [56]
3 years ago
10

2

Mathematics
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

is there any pics attached so I can help

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The length of a rectangle is six times it’s width. If the area of the rectangle is 150ft^2, find its perimeter.
sineoko [7]

Answer:

Perimeter = 70ft

Step-by-step explanation:

Area of a rectangle is:

area = (length * width)

150 = a * b

a = 6b

a = length

b = width

then:

150 = 6b*b

150 = 6b²

b² = 150/6

b² = 25

√b² = √25

b = 5ft

a = 6b

a = 6*5

a = 30ft

then:

perimeter of a rectangle is:

p = 2(a+b)

p = 2(30 + 5)

p = 2*35

p = 70ft

7 0
3 years ago
What is 8 thousands 15 hundreds 15 Tens 18 ones written in standard form ?
ryzh [129]
I'm not sure what you are trying to convey, but you cannot have more than one digit, so this is imposible
8 0
4 years ago
Read 2 more answers
Figure PQRS is rotated counterclockwise
sleet_krkn [62]
I gotchu fam, a rotation is occurring and we know this because it’s rotating 180°. Figures PQRS and P’Q’R’S’ are congruent because they are the same shape and size, just in different places.
4 0
3 years ago
15PTS PLEASE HELP ASAP!<br> (dont write random answers pls)<br> look at the pic attached:
MrMuchimi

Answer: Definition of Vertical Angles.

5 0
3 years ago
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Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
4 years ago
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