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coldgirl [10]
2 years ago
13

Trevor used the time-to-pay-off formula to calculate how many payments it

Mathematics
2 answers:
boyakko [2]2 years ago
5 0
It’s A took this test before
kirill [66]2 years ago
3 0

Answer:

The answer is D literally just took the test

Step-by-step explanation:

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David received his allowance on Sunday. He spends 1/4 of his allowance on Monday and 2/3 of the remainder on Tuesday. What part
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Answer:1/12 left

Step-by-step explanation: you should switch both into the same denominator for the fractions so that would turn into 3/12 and 8/12 then add them up to get 11/12 so he has 1/12 left

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2 years ago
Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, as shown colon
Nadusha1986 [10]

Answer:

Sadie's error is " she made error in step 2 =\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}  where a\geqslant 0"

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is 3ab\sqrt{6ab} where a\geqslant 0</h3>

Step-by-step explanation:

Given that " Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, "

It can be written as \sqrt{54a^7b^3} where a\geqslant 0

The given expression is \sqrt{54a^7b^3} where a\geqslant 0

To find Sadie's error and explain the correct answer :

Sadie's steps are

\sqrt{54a^7b^3}  where a\geqslant 0

=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}

=3ab\sqrt{6a^5b}

<h3>\sqrt{54a^7b^3}=3ab\sqrt{6a^5b} where a\geqslant 0</h3><h3><u>Now corrected steps are</u></h3>

\sqrt{54a^7b^3}  where a≥0

=\sqrt{(9\times 6)(a^{6+1})(b^{2+1})

=\sqrt{(3^2\times 6)(a^6.a^1)(b^2.b^1) (by using the identity a^{m+n}=a^m.a^n

=\sqrt{3^2\times 6\times ((a^3)^2.a)(b^2.b) (by using the identity a^{mn}=(a^m)^n )

=3ab\sqrt{6ab}

Therefore \sqrt{54a^7b^3}=3ab\sqrt{6ab}  where a\geqslant 0

<h3>The correct answer is 3ab\sqrt{6ab} where a\geqslant 0</h3>

Sadie's error is " she made error in step 2 =\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b} " where a\geqslant 0

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is 3ab\sqrt{6ab} where a\geqslant 0</h3>
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2 years ago
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