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3 years ago

How can you find discount and markup

Mathematics

1 answer:

melamori03 [73]3 years ago

5 0

By using different math operations

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Pls help photo included !!!!

AlladinOne [14]

1. 371.25

2. 742.5

3.185.625

6 0

3 years ago

What is the length of the unknown leg

likoan [24]

Answer:

6 mm

Step-by-step explanation:

Use the Pythagorean Theorem to solve for the unknown leg

$a^{2} +b^{2}=c^{2}$

Since we need to solve for a, we will manipulate the equation in terms of b and c → $a=\sqrt{c^{2}-b^{2} }$

Here, b = 7 mm and c = $\sqrt{85}$ mm

Plugging these numbers into our equation gives us $a=\sqrt{(\sqrt{85})^{2} -(7)^{2} }$ → $\sqrt{85-49}$ = $\sqrt{36}$ = 6 mm

6 0

3 years ago

What is the missing number ? <br> 1,8,27,64,125,216,?

Katena32 [7]

Answer:

Step-by-step explanation:

this is the sequence of cube. ... hence the sequence will be 1-8-27-64-125-216.

(wen 64 is already tere)

3 0

2 years ago

Evaluate.<br><br><br> 2 • {390 – [5 • (10 – 4 ÷ 2)] + 15} • 2

otez555 [7]

The correct answer is 1460

8 0

3 years ago

Complete the steps to solve the polynomial equation x3 – 21x = –20. According to the rational root theorem, which number is a po

jeka94

Answer:

Zeroes : 1, 4 and -5.

Potential roots: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ .

Step-by-step explanation:

The given equation is

$x^3-21x=-20$

It can be written as

$x^3+0x^2-21x+20=0$

Splitting the middle terms, we get

$x^3-x^2+x^2-x-20x+20=0$

$x^2(x-1)+x(x-1)-20(x-1)=0$

$(x-1)(x^2+x-20)=0$

Splitting the middle terms, we get

$(x-1)(x^2+5x-4x-20)=0$

$(x-1)(x(x+5)-4(x+5))=0$

$(x-1)(x+5)(x-4)=0$

Using zero product property, we get

$x-1=0\Rightarrow x=1$

$x-4=0\Rightarrow x=4$

$x+5=0\Rightarrow x=-5$

Therefore, the zeroes of the equation are 1, 4 and -5.

According to rational root theorem, the potential root of the polynomial are

$x=\dfrac{\text{Factor of constant}}{\text{Factor of leading coefficient}}$

Constant = 20

Factors of constant ±1, ±2, ±4, ±5, ±10, ±20.

Leading coefficient= 1

Factors of leading coefficient ±1.

Therefore, the potential root of the polynomial are $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ .

3 0

3 years ago