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3 years ago

What is the perimeter of the rectangle with a length of 23 and a width of 16?

Mathematics

1 answer:

motikmotik3 years ago

7 0

23+23+16+16=78
78 is the perimeter:)

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Select True or False for each statement. 16.32 rounded to the nearest whole number is 16 feet per second. ? 16.32 rounded to the

Maurinko [17]

Answer:

a)16.32 rounded to the nearest whole number is 16 feet per second? True

b) 16.32 rounded to the nearest tenth is 16.3.? True

c) 16.32 rounded to the nearest hundredth is 16.33. ? False

Step-by-step explanation:

Select True or False for each statement.

a) 16.32 rounded to the nearest whole number is 16 feet per second. ?

16.32 to nearest whole number = 16 feet per second. This statement above is true.

b) 16.32 rounded to the nearest tenth is 16.3. ?

The number in the tenth position is 3, the number after it is 2.

Hence, 16.32 to the nearest tenth = 16.3

Hence, the above statement is true.

c) 16.32 rounded to the nearest hundredth is 16.33. ?

The number in the hundredth position is 2. There is no number after 2.

Hence,

16.32 to the nearest hundredth is 16.32

Hence, 16.32 rounded to the nearest hundredth is 16.33 is a false statement.

5 0

2 years ago

-6y+y2+3-4y2-7+1-4y<br> Please help

vaieri [72.5K]

-6y+y2+3-4y2-7+1-4y in simplified form is -10y-3y^2-3

5 0

2 years ago

What's the opposite reciprocal of -2

Ierofanga [76]

t h e r e c i p r i c l e o f - 2 i s 2

7 0

3 years ago

Nicholas makes $2,000 per month.He spends $300 on credit card payments and $350 on an auto loan.what is his debt to income ratio

I am Lyosha [343]

Answer:

32.5%

Step-by-step explanation:

His debt payments total $300 +350 = $650.

As a fraction of his income, that is ...

$650/$2000 = 325/1000 = 32.5/100 = 32.5%

Nicholas's debt to income ratio is 32.5%.

3 0

3 years ago

Read 2 more answers

lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.

zaharov [31]

Answer with Step-by-step explanation:

Given

$f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1$

6 0

3 years ago