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3 years ago

11

The width of a rectangular computer screen is 2 inches more than its height. If the area of the screen is 323 square inches, fin

d its dimensions.

1 answer:

Lana71 [14]3 years ago

5 0

Width= 19 in
Height= 17 in

WxH=323

19x17=323

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Write an equivalent expression for 5a+30

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We can express it at 5 (a+6)

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1. Find P (phone has internet or phone plays music) using this Venn diagram.

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Step-by-step explanation:

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the venn diagram represents the results of a survey that asked participants whether they would want a rabbit or a snake as a pet

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For this case we will complete the table part by part.
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The odds of winning a contest are 3:7. What is the probability winning the contest?

DiKsa [7]

Answer:

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Step-by-step explanation:

Given: The odds of winning a contest are 3:7

To find: probability of winning the contest

Solution:

Probability refers to chances of occurrence of an event.

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So,

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3 years ago

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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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3 years ago