A store has a $120 dress. Then there is a 200% increase in the price of the dress. What is the final price of the dress?

What is the quotient (x3 – 3x2 + 3x – 2) ÷ (x2 – x + 1)?

timofeeve [1]

We need to find the quotient of the given division problem.

$\frac{x^{3}-3x^{2}+3x-2}{x^{2}-x+1}$

In order to find its quotient, we will use long division.

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

First of all, we put x in the quotient as $x^{2}$ goes into $x^{3}$ , x times.

So, we get:

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

$\text{ ..................}x^{3}-x^{2}+x$

Upon subtracting, we get:

$\text{....................}-2x^{2}+2x-2$

We can see that $x^{2}$ goes into $-2x^{2}$ , -2 times, therefore, the next term in the quotient will be -2. This makes our quotient as (x-2).

5 0

2 years ago

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Match each three-dimensional figure to its volume based on the given dimensions. (Assume π = 3.14.)

LekaFEV [45]

Answer:

The volume of the cylinder is 150.72 cm³ ⇒ last answer

The volume of the cone is 314 cm³ ⇒ 1st answer

The volume of the pyramid is 160 cm³ ⇒ 2nd answer

The volume of the pyramid is 48 cm³ ⇒ 3rd answer

Step-by-step explanation:

* Lets revise the volumes of some shapes

- The volume of the cylinder of radius r and height h is:

V = π r² h

- The volume of the cone of radius r and height h is:

V = 1/3 π r² h

- The volume of the pyramid is:

V = 1/3 × its base area × its height

* Lets solve the problem

# A cylinder with radius 4 cm and height 3 cm

∵ V = π r² h

∵ π = 3.14

∵ r = 4 cm , h = 3 cm

∴ v = 3.14 (4)² (3) = 150.72 cm³

* The volume of the cylinder is 150.72 cm³

# A cone with radius 5 cm and height 12 cm

∵ V = 1/3 π r² h

∵ π = 3.14

∵ r = 5 cm , h = 12 cm

∴ V = 1/3 (3.14) (5)² (12) = 314 cm³

* The volume of the cone is 314 cm³

# A pyramid with base area 16 cm² and height 30 cm

∵ V = 1/3 × its base area × its height

∵ The area of the base is 16 cm²

∵ The height = 30 cm

∴ V = 1/3 (16) (30) = 160 cm³

* The volume of the pyramid is 160 cm³

# A pyramid with square base of length 3 cm and height 16 cm

∵ V = 1/3 × its base area × its height

∵ The area of the square = s²

∵ The area of the base = 3² = 9 cm²

∵ The height = 16 cm

∴ V = 1/3 (9) (16) = 48 cm³

* The volume of the pyramid is 48 cm³

3 0

1 year ago

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What's 3/4 times 7/9 in simplest form

Alexxandr [17]

3/4 times 7/9:

First, we must use the rule for multiplying fraction. The rule is: a/b times c/d = ac/ bd. Let's substitute our problem into that formation.
$\frac{3 \times 7}{4 \times 9}$

Second, we can now simplify. (3 times 7 = 21) and (4 times 9 = 36).
$\frac{21}{36}$

Third, since our fraction is not in the simplest form, we can simplify it down by listing the factors of both the numerator and denominator and then finding the greatest common factor (GCF).

Factors of 21: 1, 3, 7, 21
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

The GCF is 3, since only 1 and 3 were the common factors and since 3 is the greatest common factor.

Fourth, now we can divide our numerator (21) and denominator (36) by our recently found GCF which was 3.

$21 \div 3 = 7 \\ 36 \div 3 = 12$

Our new fraction in its simplest form is 7/12.

Answer in fraction form: $\fbox {7/12}$
Answer in decimal form: $\fbox {0.5833}$

7 0

2 years ago

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Leena walked 2/3 of a mile What is 2/3 written as a sum of unit fractions with the denominator of nine

aliya0001 [1]

Bear in mind that multiplying anything, and anything whatsoever by 1, has a product of the anything itself.

so, let's multiply 2/3 by 1 then and check about,

$\bf \cfrac{same}{same}=1\qquad thus\cfrac{2}{3}\cdot 1\implies \cfrac{2}{3}\cdot \cfrac{3}{3}\implies \cfrac{6}{9}
\\\\\\
\textit{now, we can split that and write it as a sum}\implies \cfrac{1+5}{9}\implies \cfrac{1}{9}+\cfrac{5}{9}$

3 0

2 years ago

What is the horizontal asymptote for y(t) for the differential equation dy dt equals the product of 2 times y and the quantity 1

marta [7]

First, we need to solve the differential equation.
$\frac{d}{dt}\left(y\right)=2y\left(1-\frac{y}{8}\right)$
This a separable ODE. We can rewrite it like this:
$-\frac{4}{y^2-8y}{dy}=dt$
Now we integrate both sides.
$\int \:-\frac{4}{y^2-8y}dy=\int \:dt$
We get:
$\frac{1}{2}\ln \left|\frac{y-4}{4}+1\right|-\frac{1}{2}\ln \left|\frac{y-4}{4}-1\right|=t+c_1$
When we solve for y we get our solution:
$y=\frac{8e^{c_1+2t}}{e^{c_1+2t}-1}$
To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity.
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:
$$$\lim_{x\to\infty} f(x)$$=y=\frac{8e^{c_1+\infty}}{e^{c_1+\infty}-1} = 8$
When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants.
So our asymptote is at y=8.

3 0

1 year ago