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3 years ago

What is a rectangle?

Mathematics

2 answers:

enot [183]3 years ago

8 0

A rectangle is a figure with 4 straight lines, and 4 right angles. preferably with unequal adjacent sides.

Hunter-Best [27]3 years ago

6 0

A shape with 4 corners, that is longer on two opposite sides, and shorter on the other two sides

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Somebody help me with this?

trasher [3.6K]

Answer:

Zero's

Step-by-step explanation:

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2 years ago

If I have more 15 Pokemon cards than Austin, which expression shows how many | have?

Nataly_w [17]

Answer:

a+15=x

Step-by-step explanation:

Austin is represented as variable a, and you have 15 more cards.

x is your amount of cards

5 0

2 years ago

30 points Help ASAP How do division properties help you evaluate expressions? Drag a word or number to the box to correctly comp

yKpoI14uk [10]

Answer:

Numerator , 0

Anything divided by 0 is always 0

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3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

2 years ago

Read 2 more answers

$12.50 for 5 ounces

iragen [17]

Answer:

45 copies per minute

Look at drawing for an explanation

6 0

3 years ago