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3 years ago

Find the fractional side lengths of a rectangle that has a perimeter of 64 5/6 inches.

Mathematics

1 answer:

Margarita [4]3 years ago

8 0

H oh oh oh ola oholglvl k,

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alukav5142 [94]

I think the answer would be 4/5

8 0

2 years ago

Need help with this problem

insens350 [35]

That equation just equals ln(e^(-4/3)) which can be simplified to -4/3 because ln and e cancel each other out only leaving the power

4 0

3 years ago

5x-60 = -60 +2x +3x how many solutions does the equation have

xeze [42]

Answer:

$infinitely \: many \: solution$

Step-by-step explanation:

1) Cancle -60 on both sides.

$5x = 2x + 3x$

2) Simplify 2x + 3x to 5x.

$5x = 5x$

3) Since both sides are equal, there are infinitely many solutions.

$infinietly \: many \: solution$

Therefor, this equation has infinitely many solutions.

5 0

3 years ago

Tell someone a story about "1 more and then 1 more

gayaneshka [121]

Once upon a time there was a guy named Jeff. he went to a new school where he moved. so he was all by himself with no friends at all.. so then he met this guy named Zac so it was 1+1=2. and they were bff so they lived happily ever after together being bffs

4 0

3 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

2 years ago