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3 years ago

10

Three single-digit numbers multiply to make 168

1 answer:

sleet_krkn [62]3 years ago

4 0

3, 8 and 7 because if you multiply 3 by 8 you get 24 and if you multiply 24 by 7 you
get 168

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Paraphin [41]

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3 years ago

How do you factor this polynomial as a product of more than two factors?<br> x3+8x2+15x

umka21 [38]

$x^3+8x^2+15x =\\
x(x^2+8x+15)=\\
x(x^2+3x+5x+15)=\\
x(x(x+3)+5(x+3))=\\
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3 years ago

What is the value of b in the equation below? 9^-8 x 9^-2 = a^b

vfiekz [6]

Answer:

b= -10

Step-by-step explanation:

9^ -8 * 9^ -2

We know that a^b * a^c = a^ (b+c)

9^ (-8+-2)

9^ -10

We want it in the form a^b

a=9

b= -10

7 0

3 years ago

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almond37 [142]

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7 0

3 years ago

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago