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3 years ago

I'm stuck, please help me

Mathematics

2 answers:

Sav [38]3 years ago

3 0

60/48
x100
60/48=1.25
1.25x100
125%

uranmaximum [27]3 years ago

3 0

Answer:

The answer is 80%

Step-by-step explanation:

First you divide 48 by 60 and then multiply it by 100 and you get 80.

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nirvana33 [79]

Answer:

Yes,she had enough

½pound=8ounces

she needs a total of 24ounces

Step-by-step explanation:

She had enough because ½pound is 8ounces thus 16ounces+8ounces add up to 24ounces.

7 0

4 years ago

Help me on this math problem

Anon25 [30]

Answer:

D: t = u/9 + 7 (that is simplified all of the way) for your case it is t = u + 63/9 so D

Explanation:

Just ask if you need an explanation. Also, does anyone do LD debate?

4 0

3 years ago

(ASAP) Solve the system by elimination <br><br> -2x+2y+3z=0<br> -2x-y+z=-3<br> 2x+3y+3z=5

Fantom [35]

To solve by elimination you need to solve for each variable, in this case x, y, and z
x = 1/3
y = 7/3
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7 0

4 years ago

Read 2 more answers

Without a calculator how would you solve this?

mina [271]

Answer:

the largest number that is less than (2+√3)^6 is 2701

Step-by-step explanation:

$\bigstar \ (2+\sqrt{3})^6= \\\\C{}^{0}_{6}2^{6}+C{}^{1}_{6}2^{5}\sqrt{3}^1+C{}^{2}_{6}2^{4}\sqrt{3}^2+C{}^{3}_{6}2^{3}\sqrt{3}^3+C{}^{4}_{6}2^{2}\sqrt{3}^4+C{}^{5}_{6}2^{1}\sqrt{3}^5+C{}^{6}_{6}\sqrt{3}^6$

$=64+192\sqrt{3}+720 +480\sqrt{3} +540+108\sqrt{3} +27\\=1351+780\sqrt{3}$

$\bigstar\ (2-\sqrt{3} )^6 =\\\\1351-780\sqrt{3}$

$\bigstar\ \bigstar\ (2+\sqrt{3} )^6+(2-\sqrt{3} )^6 =\\\\1351+780\sqrt{3}+1351-780\sqrt{3}$

$\Longrightarrow(2+\sqrt{3} )^6+(2-\sqrt{3} )^6 =2702$

$\Longrightarrow(2+\sqrt{3} )^6 =2702-(2-\sqrt{3} )^6$

$0 < \left( 2-\sqrt{3} \right) < 1 \Longrightarrow 0 < \left( 2-\sqrt{3} \right)^{6} < 1 \Longrightarrow-1 < \left( 2-\sqrt{3} \right)^{6} < 0$

$\Longrightarrow2702-1 < 2702-\left( 2-\sqrt{3} \right)^{6} < 2702+0$

$\Longrightarrow 2701 < \left( 2+\sqrt{3} \right)^{6} < 2702$

8 0

2 years ago

PLEASE HELP ME OUT AND NOT LET A BOT ANSWER THIS :(

laiz [17]

Answer: There is no such thing. All number have to be rational or irrational

8 0

3 years ago

I'm stuck, please help me​

I'm stuck, please help me