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2 years ago

Classify the angle as acute, right, obtuse, or straight.

Mathematics

1 answer:

taurus [48]2 years ago

5 0

Answer:

Step-by-step explanation:

Our eyes tell us that the angle is less than 90 degrees. That's the definition of Acute.

A right angle = 90 degrees.

An obtuse angle is larger than 90 degrees.

A straight angle is a line which = 180 degrees.

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Solve for x, assuming a, b, and c are negative constants.<br><br> (ax + b)/c ≤ b

Rudik [331]

(ax + b)/c ≤ b

ax+b ≤ cb

ax ≤ cb - b

x ≤ (cb-b)/a

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3 years ago

An $8 item is marked up 75%. What is the cost of the item?

UNO [17]

14 dollars because it’s only 75 % out of. 8

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2 years ago

you have already saved $55. you earn $9 per hour at your job. you are saving for a bicycle that cists $199. what inequality repr

Anon25 [30]

(199-55)/9= t
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3 years ago

Read 2 more answers

A website received 2500 registered members in July. And the number of registered members in August decreased to 2245. What was t

quester [9]

Number of registrations a website received during July = 2500

Number of registrations a website received during August = 2245

Number by which the registrations decreased :

$=\tt 2500 - 2245$

$=\tt 255$

Thus, the registrations decreased by 255.

Let x be the perrcentage of decrease.

Which means :

$=\tt 255 = \frac{x}{100} \: of \: 2500$

$= \tt255 = \frac{x \times 2500}{100}$

$=\tt 255 = \frac{2500x}{100}$

$= \tt255 \times 100 = 2500x$

$= \tt25500 = 2500x$

$=\tt \frac{25000}{2500} = x$

$\color{plum} =\tt \bold{10.2\%}$

▪︎Thus, the percentage of decrease in the number of registered members = 10.2%

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2 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

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3 years ago