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3 years ago

1. If the original price is $75 and you get a 30% discount. What is the sale price?

Mathematics

2 answers:

Flura [38]3 years ago

7 0

Answer:

the sale price is 52.50

Step-by-step explanation:

30% discount means that the person pays 70% of the original price

multiply 75 by 70% written as a decimal 0.70

75*0.70 = 52.50

Alex Ar [27]3 years ago

7 0

Answer:
The 30% discount is $22.50 off the original price. Thus, a 30% discount on $75 brings the total to $52.50.

Explanation:
First, calculate the 30% discount.

75 • .30 = 22.5

So, this means the 30% discount gives us $22.50 off the original price.

Next, subtract the discount from the original price to find the final total.

75 - 22.5 = 52.5

So, the final total is $52.50.

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Find the first five terms of the sequence defined below, where n represents the position of a

SOVA2 [1]

Answer:

Your answer is going to be 4, 7, 10, 13, 16

Step-by-step explanation:

n=1, 3*1+1=4

n=2, 3*2+1=7

n=3, 3*3+1=10

n=4, 3*4+1=13

n=5, 3*5+1=16

6 0

3 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

Help anyone ?????????

elena-14-01-66 [18.8K]

Ans: -4.3

Integers are all whole numbers negative, positive, and 0

Hope this helped :)

7 0

4 years ago

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puteri [66]

The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

<h3>How to prove an Isosceles Triangle?</h3>

Let ABC be an isosceles triangle such that AB = AC.

Let AD be the bisector of ∠A.

We want to prove that BD=DC

In △ABD & △ACD

AB = AC(Thus, △ABC is an isosceles triangle)

∠BAD =∠CAD(Because AD is the bisector of ∠A)

AD = AD(Common sides)

By SAS Congruency, we have;

△ABD ≅ △ACD

By corresponding parts of congruent triangles, we can say that; BD=DC

Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

Read more about Isosceles Triangle at; brainly.com/question/1475130

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2 years ago

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Lapatulllka [165]

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3 years ago

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