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adelina 88 [10]

3 years ago

9

HELP FAST!!! Here are two rectangles. The length and width of one rectangle are 8 and 5. The width of the other rectangle is 5,

but its length is unknown so we labeled it x. Write an expression for the sum of the areas of the two rectangles.

1 answer:

deff fn [24]3 years ago

6 0

Answer:

5•x=? and 8•5=40, personal opinion: the left rectangle is most likely a square so if you need to solve it it's 5•5=10

Step-by-step explanation:

A(area)=l (length)•w (width)

just plug in your numbers, in this case it would be 8(length) times 5(width)

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Dvinal [7]

Question 11)

Answer:

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Step-by-step explanation:

Given the expression

$\log _2\left(63\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(\frac{63}{9}\right)$

$=\log _2\left(\frac{63}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{63}{9}=7$

$=\log _2\left(7\right)$

$=2.80$

Therefore,

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Question 12)

Answer:

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

Step-by-step explanation:

Given the expression

$\log _2\left(3\right)+\log _2\left(15\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _2\left(3\right)+\log _2\left(15\right)=\log _2\left(3\cdot \:15\right)$

$=\log _2\left(3\cdot \:15\right)-\log _2\left(9\right)$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:15=45$

$=\log _2\left(45\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(45\right)-\log _2\left(9\right)=\log _2\left(\frac{45}{9}\right)$

$=\log _2\left(\frac{45}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{45}{9}=5$

$=\log _2\left(5\right)$

$=2.32$

Therefore,

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

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