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3 years ago

Please help, i need this today!

Mathematics

1 answer:

pav-90 [236]3 years ago

3 0

The answer is C because the x value doesn’t repeat

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Sam opened a savings account with an initial deposit of $2,000. Since then, he has never made any other deposits or withdrawals.

notsponge [240]

If my math is correct B or C is what it comes down to

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2 years ago

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What is the number 5 2/7 written as a improper fraction

WARRIOR [948]

Answer:

the answer is 37/7

Step-by-step explanation:

u multiply the 5 and the seven and add the 2 and u get 37/7 so u keep the denominator the same

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2 years ago

What is the sum of 1+3+5+7+...1209

Natalka [10]

the sum of 1+3+5+7+1209 is.....1225

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2 years ago

11 more than five of a certain number is twenty more than 2 times of that number. what is the number?

RUDIKE [14]

Answer:

Let the number be x

Given:

11 + 5x = 20 + 2x

5x - 2x = 20 - 11

3x = 9

x = 3

The number is 3.

Step-by-step explanation:

6 0

3 years ago

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What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$

6 0

3 years ago