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3 years ago

6

Maxine spent 15 hours doing her homework last week. This week she spent 18 hours doing homework. She says that she spent 120% mo

re time doing homework this week. Is she correct? Show your work to justify your decision. (5 points)

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

she is correct beacuse if she worked more harder this week than last week it shows beacuse she did 15 hours

and she did 18 hours as well

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Simplify 2a^2b^3(4a^2+3ab^2-ab)

-BARSIC- [3]

2a^2b^3(4a^2+3ab^2-ab)=?
<span>
is what I presume you actually meant. </span>
<span>
Pull out the common factors of (4a^2+3ab^2-ab) and you will get </span>
<span>
a(4a+3b^2 -b) </span>

Substitute this back into the original equation and you get
<span>
2a^2b^3[a(4a+3b^2-b)] = </span>
2a^3b^3(4a+3b^2-b) =
<span>2a^3b^3(4a-b+3b^2)
</span>

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$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest e

ch4aika [34]

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = $p_1$ = $A

And The money invested at 6% = $p_2$ = $13957 - $A

Let The interest earn at 7% = $I_1$

And The interest earn at 6% = $I_2$

$I_1$ - $I_2$ = $833.73

Let The time period = 1 year

Now,<u> From Simple Interest method</u>

Simple Interest = $\dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}$

Or, $I_1$ = $\dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}$

Or, $I_1$ = $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$

And

$I_2$ = $\dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}$

Or, $I_2$ = $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$

∵ $I_1$ - $I_2$ = $833.73

So, $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$ - $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$ = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = $\dfrac{167115}{13}$

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102 . Answer

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