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3 years ago

How can polygons be considered a subcategory of two-dimensional figures?

Mathematics

1 answer:

Helga [31]3 years ago

4 0

Yes; for example triangles, squares, and rectangle.

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Convert to radical form (3) 2/3

UNO [17]

Answer:

The radical form of 3 2/3 is

$\sqrt[3]{3 {}^{2} }$

3 0

3 years ago

A number divided by 43 has a quotient of 3 with 28 as a remainder. Find the number. Show your work. Write another division probl

antiseptic1488 [7]

A quotient of 3 with 28 as a remainder means that 43 fits inside our number 3 times, and you have 28 more units as a remainder.

So, our number is

$3\times 43 + 28 = 129+28=157$

To write another division problem that has a quotient of 3 and a remainder of 28, we simply choose another number to substitute 43 in the expression above. For example, if we choose 100, the expression becomes

$3\times 100 + 28 = 300+28=328$

Which means that 328 has quotient of 3 with 28 as a remainder when divided by 100.

4 0

3 years ago

a sports store sells three different packs of tennis balls. which pack is the best to buy? 4 sleeves for 11.49, 6 sleeves for 16

PIT_PIT [208]

Answer:

9 sleeves for 22.99

Step-by-step explanation

The first pack has a unit rate of $11.49/ 4 balls

=$2.8725

The second pack has a unit rate of $16.79/6balls

=$2.798 per ball

the third pack has a unit rate of $22.99/9balls

=2.554

So, I would definitely go for the third pack since it has the least cost.

6 0

2 years ago

Write an absolute value inequality that

fiasKO [112]

Answer:

All real numbers greater than or equal to -1 and less than or equal to 2

Step-by-step explanation:

Let

A_1 ------> the area of rectangle

A_2 ----> the area of triangle

we know that

$\left|A_1-A_2\right|\le 6$

Find out the area of rectangle

$A_1=4(x+2)=(4x+8)\ units^2$

Find out the area of triangle

$A_2=(1/2)(4)(5)=10\ units^2$

substitute

$\left|(4x+8)-10\right|\le 6$

Simplify

$\left|4x-2\right|\le 6$

First case (positive value)

$+(4x-2) \leq 6$

$4x \leq 6+2$

$4x \leq 8$

$x \leq 2$

Second case (negative value)

$-(4x-2) \leq 6$

Multiply by -1 both sides

$(4x-2) \geq -6$

$(4x \geq -6+2$

$4x \geq -4$

$x \geq -1$

therefore

The solution for x is the interval -----> [-1,2]

All real numbers greater than or equal to -1 and less than or equal to 2

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3 years ago

OK Rolls Rejected Rolls Rolls to Repair

Aliun [14]

I think it is A. 78.9

8 0

3 years ago

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