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2 years ago

Find the area of ABCHelp

Mathematics

1 answer:

Katarina [22]2 years ago

6 0

The value of the scale factor is 3. Then the height and area of the triangle ΔABC will be 9 ft and 54 square ft.

<h3>What is dilation?</h3>

Dilation means the changing of the size of the object without changing the shape. The size of the object may be increased or decreased based on the scale factor.

The area of the triangle ΔDEF is 6 square ft. And the base length of the triangle ΔDEF is 4 ft.

The area of the triangle is given as

$\rm Area = \dfrac{1}{2} \times bh\\\\$

Then the height of the triangle ΔDEF will be

$\rm 6 = \dfrac{1}{2} \times 4 \times h\\\\h = \dfrac{12}{4}\\\\h = 3$

Then the scale factor will be

$\rm SF = \dfrac{12}{4} = 3$

Then the height of the triangle ΔABC will be

$\rm \dfrac{H}{3} = 3\\\\H \ = 9$

Then the area of the dilated triangle ΔABC will be

$\rm Dilated \ Area = \dfrac{1}{2} \times 12 \times 9\\\\Dilated \ Area = 54 \ ft^2$

More about the dilation link is given below.

brainly.com/question/2856466

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A manufacturing company with 450 employees begins a new product line and must increase their number buy 18% how many total emplo

Soloha48 [4]

Answer:

After increasing their employees by 18% they will now have 531 total employees.

Step-by-step explanation:

0.18 x 450 = 81

81 + 450 = 531

7 0

3 years ago

Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is

ohaa [14]

The vectors in $S$ form a basis of $P_2$ if they are mutually linearly independent and span $P_2$ .

To check for independence, we can compute the Wronskian determinant:

$\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0$

The determinant is non-zero, so the vectors are indeed independent.

To check if they span $P_2$ , you need to show that any vector in $P_2$ can be expressed as a linear combination of the vectors in $S$ . We can write an arbitrary vector in $P_2$ as

$p=ax^2+bx+c$

Then we need to show that there is always some choice of scalars $k_1,k_2,k_3$ such that

$k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p$

This is equivalent to solving

$(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c$

or the system (in matrix form)

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

This has a solution if the coefficient matrix on the left is invertible. It is, because

$\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0$

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}$

Then

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}$

$\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}$

A solution exists for any choice of $a,b,c$ , so the vectors in $S$ indeed span $P_2$ .

The vectors in $S$ are independent and span $P_2$ , so $S$ forms a basis of $P_2$ .

5 0

3 years ago

More Algebra hw-- :/

grin007 [14]

Answer:
-11x=-66
x=6

Explanation:
-11x-12=-78
1) Add 12 to both sides.
-11x=-78+12
-11x=-66
2) Divide both sides by -11.
x=-66/-11
x=6

Hope this helps! Please leave a thanks and brainliest answer if you can :)

3 0

2 years ago

Line l1 passes through (-2, 5), and (-1, -10)

zalisa [80]

The equations of the lines l1 with points (-2, 5), and (-1, -10) and the line l2 with points (5, 15), and (3, 8), gives the coordinate of the intersection between the lines as the point; (-45/37, -250/37)

<h3>Which method can be used to describe the lines to find the intersection point?</h3>

The slope, m1, of line 1 l1 is found as follows;

m1 = (5 - (-10))/(-2- (-1)) = -15

The equation of line l1 in point and slope form is therefore;

y1 - 5 = -15•(x - (-2))

Which gives;

y1 = -15•(x - (-2)) + 5 = -15•x - 25

y1 = -15•x - 25

The slope, m2, of line 2 l2 is found as follows;

m2 = (15 - 8)/(5 - 3) = 3.5

Equation of line l2 is therefore;

y2 - 15 = 3.5•(x - 5)

Which gives;

y2 = 3.5•(x - 5) + 15 = 3.5•x - 2.5

y2 = 3.5•x - 2.5

At the intersection point, we have;

y1 = y2

Therefore;

-15•x - 25 = 3.5•x - 2.5

18.5•x = -22.5

x = -22.5/18.5 = -45/37

y2 = 3.5•x - 2.5

At the intersection point, we have;

y = y2 = 3.5×(-22.5/18.5) - 2.5 = -250/37

y = -250/37

The coordinates of the intersection between the lines is therefore;

(-45/37, -250/37)

Learn more about the equation of a straight line here:

brainly.com/question/13763238

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3 0

1 year ago

Student tickets to the school football game cost $5.80 each and general admission

Tems11 [23]

Answer:

x + y = 380

5.8x + 7.4y = 2372

Step-by-step explanation:

7 0

3 years ago