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3 years ago

Which of these statements is not one of Euclid axioms

Mathematics

2 answers:

earnstyle [38]3 years ago

8 0

Answer:

Things which are equal to the same thing are also equal to one another. If equals be added to equals, the wholes are equal. If equals be subtracted from equals, the remainders are equal. Things which coincide with one another are equal to one another.

lorasvet [3.4K]3 years ago

6 0

They are a equally big fan of the best way out here and the cashapp method of a and a full of flavor that shi was the best way we could have seen in a while and it is

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What is the length of the hypotenuse of the trim angle ?

Ksju [112]

So for this, you'll be using the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ . In this case, 10 and 24 are the legs, and AB is the hypotenuse. Our equation will be $10^2+24^2=AB^2$

Firstly, solve the exponents: $100+576=AB^2$

Next, combine 100 and 576: $676=AB^2$

And lastly, square root both sides to get $26=AB$

And in short, Line AB is 26 ft, or the first option.

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3 years ago

A trapezoid has four sides with two sides that are the same length. Let the two sides have length x. The longest side has a leng

Oduvanchick [21]

Answer:

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Step-by-step explanation:

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3 years ago

Which of the following statements would be supported by both a communist and a utopian socialist?

guajiro [1.7K]

Answer:

Step-by-step explanation:

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3 years ago

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Select the statement that best justifies the conclusion based on the given information.

AURORKA [14]

I think it would be substitution

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3 years ago

Read 2 more answers

What is x+4/5=2/3- x-3/6

natima [27]

$x+\frac{4}{5}=\frac{2}{3}-x-\frac{\not3^1}{\not6_2}\\\\x+\frac{4}{5}=\frac{2}{3}-x-\frac{1}{2}\\\\find\ LCD(\frac{4}{5};\ \frac{2}{3};\ \frac{1}{2})=2\times3\times5=30\\\\multiply\ both\ sides\ by\ LCD=30\\\\30x+30\cdot\frac{4}{5}=30\cdot\frac{2}{3}-30x-30\cdot\frac{1}{2}\\\\30x+6\cdot4=10\cdot2-30x-15\cdot1\\\\30x+24=20-30x-15$

$30x+24=5-30x\ \ \ \ |subtract\ 24\ from\ both\ sides\\\\30x=-19-30x\ \ \ \ \ |add\ 30x\ to\ both\ sides\\\\60x=-19\ \ \ \ \ |divide\ both\ sides\ by\ 60\\\\\boxed{x=-\frac{19}{60}}$

$\frac{x+4}{5}=\frac{2}{3}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{2\cdot2}{3\cdot2}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4}{6}-\frac{x-3}{6}\\\\\frac{x+4}{5}=\frac{4-(x-3)}{6}\\\\\frac{x+4}{5}=\frac{4-x+3}{6}\\\\\frac{x+4}{5}=\frac{7-x}{6}\ \ \ \ |cross\ multiply$

$6(x+4)=5(7-x)\\\\6(x)+6(4)=5(7)-5(x)\\\\6x+24=35-5x\ \ \ \ \ |subtract\ 24\ from\ both\ sides\\\\6x=11-5x\ \ \ \ \ |add\ 5x\ to\ both\ sides\\\\11x=11\ \ \ \ \ |divide\ both\ sides\ by\ 11\\\\\boxed{x=1}$

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3 years ago