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3 years ago

Write a polynomial in standard form that represents the area of the shaded region.

Mathematics

2 answers:

trapecia [35]3 years ago

7 0

Answer:

p^2 - 2p - 3.

Step-by-step explanation:

The area of a triangle = 1/2 * base * height so here we have

Area = 1/2 (p + 1)(2p - 6)

= 1/2( 2p^2 - 6p + 2p - 6)

= 1/2 ( 2p^2 - 4p - 6)

= p^2 - 2p - 3.

Kamila [148]3 years ago

4 0

1/2•(2p-6)(p+1)
1/2•(2p^2+2p-6p-6)
1/2•(2p^2-4p-6)
p^2-2p-3

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A mail order company charges 4.5% for shipping and handling. If the total order is 54.34 how much was the order total before shi

Andrews [41]

Answer:

$52

Step-by-step explanation:

Given that :

Amount charged for shipping and handling = 4.5%

Let cost of order = x

Total amount charged = 54.34

However,

total. Amount charged = (total cost + handling and shipping)

54.34 = x + 0.045x

54.34 = 1.045x

Order total before shipping = 54.34 / 1.045

Order total before shipping = 52

6 0

3 years ago

Suppose you have a right triangle with congruent legs and a hypotenuse that measure (12sqrt(5))/5 What is the length of the smal

anastassius [24]

The length of the smaller leg is 3.79

<h3>How to determine the length of the smaller leg?</h3>

Represent the smaller leg with x.

So, we have:

$x^2 + x^2 = ((12\sqrt5)/5)^2$ -- Pythagoras theorem

This gives

2x^2 = 144/5

Divide by 2

x^2 = 72/5

This gives

x^2 = 14.4

Take the square root

x = 3.79

Hence, the length of the smaller leg is 3.79

Read more about right triangles at

brainly.com/question/6322314

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4 0

2 years ago

The center of the inscribed circle of Triangle ABC is point , and the center of the circumscribed circle of Triagnle ABC is poin

quester [9]

Answer:

Part A: The center of The inscribed circle is the point S

Part B: The center of The circumscribed circle is the point P

Step-by-step explanation:

Part A: Find the center of The inscribed circle of ΔABC

The inscribed circle will touch each of the three sides of the triangle in exactly one point,

The center of the inscribed circle is the point of intersection between the angle bisectors of the triangle.

So, According to the previous definition:

A₁A₂ and B₁B₂ are the angle bisectors of ∠A and ∠B

So, the inter section between them is the center of The inscribed circle

So, the center of The inscribed circle is the point S

=====================================================

Part B: Find the center of The circumscribed circle of ΔABC

The circumscribed circle is the circle that passes through all three vertices of the triangle.

The center of the circumscribed circle is the point of intersection between the perpendicular bisectors of the sides.

So, According to the previous definition:

P₁P₂ and Q₁Q₂ are the perpendicular bisectors of AB and BC

So, the inter section between them is the center of The circumscribed circle

So, the center of The circumscribed circle is the point P

5 0

4 years ago

triangle MNO is an equilateral triangle with sides measuring 16 units What is the height of the triangle?

fiasKO [112]

see the attached figure to better understand the problem

we know that

The equilateral triangle has three equal sides

in the equilateral triangle ABC

$AB=BC=AC=16 units$

the height of the triangle is the segment BD

in the right triangle BCD

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

solve for BD

$BD^{2}=BC^{2}-DC^{2}$

substitute the values

$BD^{2}=16^{2}-8^{2}$

$BD^{2}=192$

$BD=\sqrt{192}\ units$

therefore

the answer is

the height of the triangle is $\sqrt{192}\ units$

7 0

3 years ago

Read 2 more answers

Please help will give brainliest

allsm [11]

Answer: it's equal

Step-by-step explanation:

6 0

3 years ago