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Dmitry [639]
2 years ago
6

Calculate the area of a triangle ABC with altitude CD, given A (-3, -4), B (-6, 2), C (0, 0), and D (-4, -2). A. 14 square units

b. 15 square units c. 18 square units d. 20 square units
Mathematics
2 answers:
kobusy [5.1K]2 years ago
8 0

Answer:

i used an online caluculator and its 15

Step-by-step explanation:

lina2011 [118]2 years ago
5 0

Answer:

15

Step-by-step explanation:

took the test

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How do you make a parabola?
sleet_krkn [62]
Make an equation where

Y =
(any number not zero) times X² + (any number at all) times X + (any number at all)

The graph of that equation is guaranteed to be a parabola.
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3 years ago
A) a shopkeeper buys a camera for $300 and sells it at $360. Calculate the percentage profit.
Lilit [14]
A) profit/original price x100 =percentage profit

(Profit: 360-300=$60)

=60/300 x100
=20%

b) two cameras (original price): 300x2= $600
two cameras (price sold): 360x2 = $720

Profit without discount: 720-600= $120

120-100= $20 discount

20/720 x100 =2.78%
4 0
2 years ago
Read 2 more answers
A bottle of medicine contains 40 doses. How many doses are in 3 1/3 bottles?
sdas [7]
Hi there! So, there are 40 doses in 1 bottle. To find out the number of doses in 3 1/3 bottles, we have to multiply both numbers together. 3 1/3 as an improper faction is 10/3. 40/1 * 10/3 is 400/3 or 133 1/3 in simplest form. There are 133 1/3 doses of medicine in 3 1/3 bottles.
7 0
3 years ago
X + 4 is prime x2 – 9 can be factored using the formula.
Anna71 [15]

We are given two binomials: x+4 , x^2-9.

x+4 can't be factored. Therefore,  it is a prime.

Let us work on x^2-9.

9 could be written as 3^2.

Therefore, x^2-9 = x^2 - 3^2.

Now, we can apply difference of the squares formula to factor it.

We know a^2 -b^2 = (a-b) (a+b).

Therefore,  x^2 - 3^2 can be factored as (x-3) (x+3).

So, x^2-9 is not a prime binomial because it can be factored as (x-3) (x+3).

3 0
3 years ago
Read 2 more answers
Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
2 years ago
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