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3 years ago

In a rectangle, how do you determine what is length and what is width?

Mathematics

1 answer:

Ann [662]3 years ago

6 0

Going side to side(horizontal) is always length and going up and down (vertical) width

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What is the equation of the line ?

xxTIMURxx [149]

Y=-1/3x+5 is the correct answer:)

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3 years ago

Combine the like terms to create an equivalent expression. 7k-k+19

Harlamova29_29 [7]

Answer:

6k +19

Step-by-step explanation:

7k-k+19

Combine like terms

7k-k = 6k

The expression becomes 6k +19

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3 years ago

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The measure of angle BQS is 78 degrees. What is m angle NQS? M angle NQS=___

Elena L [17]

I think your answer is 30

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3 years ago

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Nora's paper airplane flew 9 1/6 feet. That was 2 3/4 feet farther than Max's plane flew. Write and solve an equation to show ho

stepan [7]

Answer:

Max's paper airplane flew $\frac{77}{12}$ feet or $6\frac{5}{12}$ feet.

Step-by-step explanation:

As Nora's paper airplane flew 9 1/6 feet

i.e.

$9\frac{1}{6}=\frac{55}{6}$

Nora's paper was farther than Max's plane flew by = 2 3/4 feet

i.e.

$2\frac{3}{4}=\frac{11}{4}$

Thus the equation to show how far Max's paper airplane flew is:

$\:\:9\frac{1}{6}-2\frac{3}{4}$

$=\frac{55}{6}-\frac{11}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}6,\:4:\quad 12$

$\mathrm{Adjust\:Fractions\:based\:on\:the\:LCM}$

$=\frac{110}{12}-\frac{33}{12}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{110-33}{12}$

$\mathrm{Subtract\:the\:numbers:}\:110-33=77$

$=\frac{77}{12}$

$=6\frac{5}{12}$

Therefore, Max's paper airplane flew $\frac{77}{12}$ feet or $6\frac{5}{12}$ feet.

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3 years ago

Find how many solutions there are to the given equation that satisfy the given condition. X1 + x2 + x3 = 22, each x; is a positi

DiKsa [7]

Answer:

Step-by-step explanation:

Given that there are three variables satisfying the equation

$X1 + x2 + x3 = 22$

Here each x is given to be a positive integer

i.e. solution set for each of the variable can be any integer from 1 to 20 at most.(because if two other integers are 1 each third has to be 20)

Hence solution set can be of the form

$(x1,x2,x3) =(1,1,22) (1,2,21) (1,3,20).....$

$=(2,1,19) (2,2,18),...\\=(3,1,18) (3,2,17),....\\...\\...\\=(20,1,1)$

If x1 =1, there are 20 solution sets

If x1 =2,there are 19

...

If x1 =20 there is 1 set

Hence total solutions can be $= 20+19+...+1\\=\frac{20(21)}{2} =210$

6 0

3 years ago

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