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3 years ago

Help me please i dont know how to do all of this ill really appreciate it please

Mathematics

2 answers:

kherson [118]3 years ago

8 0

Okay. First things first, we will distribute the -7 out to the numbers in the parenthesis. To do that, you will multiply each number in there by -7. -7 * -19c is 133 and -7 * -6 is 42. That makes the numbers become 133c + 42. We will combine like terms with -14c. 133c - 14c is 119c. The expression becomes 119c + 42. The expression in simplest form is 119c + 42.

anzhelika [568]3 years ago

4 0

Hello!

The expression: $-14c-7(-19c-6)$ .

First step: Apply the Distributive Property; it is $a(b + c) = ab + ac$ .

$-14c-7(-19c-6)$
$-14c+133c+42$

[A negative times a negative equals a positive.]
Next step: Combine like terms.

$-14c+133c+42$
$119c+42$

The simplified expression is $119c+42$ .

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

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Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

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Required

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$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

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$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

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$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

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$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

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$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

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$(a - 1)^2 = 0$

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$b + 1 - 1 = 0 - 1$

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$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

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Step-by-step explanation:

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There addition, but there IS subtraction.

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7 is what is left over.

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