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denpristay [2]
3 years ago
8

Triangle A’ B’ C’ is a dilation of triangle abc what is the scale factor

Mathematics
2 answers:
Ksivusya [100]3 years ago
7 0

Answer:

1/2

Step-by-step explanation:

AC=18 and A'C'=9

AB(x)=A'B'

18(x)=9

x=9/18

x=1/2

Anon25 [30]3 years ago
4 0
<h2>Answer:</h2>

The scale factor is:

                          1/2

<h2>Step-by-step explanation:</h2>

We know that the scale factor of a dilation is the ratio of the length of the dilated i.e. transformed figure to the corresponding lengths of the original figure.

Here in the original triangle i.e. ΔABC the length of the side AC is: AC=18 units

and the corresponding side of the dilated triangle i.e. ΔA'B'C' is: A'C'= 9 units

Hence, the scale factor is given by:

\text{Scale factor}=\dfrac{\text{Length of A'C'}}{\text{Length of AC}}\\\\i.e.\\\\\text{Scale factor}=\dfrac{9}{18}\\\\\text{Scale factor}=\dfrac{1}{2}

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Step-by-step explanation:


8 0
3 years ago
In the figure shown, BC is parallel to
arlik [135]

Answer:

52 units.

Step-by-step explanation:

If we drop a perpendicular line from point C to AD and call the point ( on AD) E we have a right triangle CED.

Now CE = 6 and as the whole figure is symmetrical about the dashed line,

ED = (26 - 10)/2

= 8.

So by Pythagoras:

CD^2 = 6^2 + 8^2 = 100

CD = 10.

So, as AB = CD,

the perimeter = 10 + 26 + 2(8)

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5 0
2 years ago
Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.
Mama L [17]

Multiplying both sides by y^2 gives

xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1

so that substituting v=y^3 and hence \frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx} gives the linear ODE,

\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1

Now multiply both sides by 3x^2 to get

x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2

so that the left side condenses into the derivative of a product.

\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2

Integrate both sides, then solve for v, then for y:

x^3v=\displaystyle\int3x^2\,\mathrm dx

x^3v=x^3+C

v=1+\dfrac C{x^3}

y^3=1+\dfrac C{x^3}

\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}

6 0
3 years ago
Solving a word problem using a quadratic equation with rationa...
choli [55]

Answer:

length : 8 m

width : 13/2 = 6.5 m

Step-by-step explanation:

area of rectangle = l(ength)×w(idth) = 52

l = 2w - 5

=> (2w - 5)×w = 52

=> 2w² -5w - 52 = 0

we know, when transforming this quadratic equation into a multiplication of two expressions of w, that we need to have 2w and w (to make 2w²), and then which 2 numbers multiply to 52 and add to -5 (when one is multiplied by 2 due to 2w) : -13 and 4

=> (2w - 13)×(w + 4) = 0

=> 2 solutions for w :

2w-13 = 0

2w = 13, w = 13/2

w+4 = 0, w = -4

negative values for a side length of a shape do not make sense, so the only usable solution is w = 13/2

=> l = 2w - 5 = 13 - 5 = 8

8 0
3 years ago
A group of friends wants to go to the amusement park. They have $94.25 to spend on parking and admission. Parking is $14.75, and
Korvikt [17]

Answer: 6

Step-by-step explanation:

From the question, we are informed that a group of friends wants to go to the amusement park and that they have $94.25 to spend on parking and admission.

We are further told that Parking is $14.75, and tickets cost $13.25 per person, including tax. The equation that can be used to solve the number of people who can go to the amusement park would be:

= 14.75 + 13.25p

Since they've $94.25 to spend, the number of people that can go to the amusement park will be:

14.75 + 13.25p = 94.25

13.25p = 94.25 - 14.75

13.25p = 79.50

p = 79.50 / 13.25

p = 6

6 people can go to the amusement park

3 0
2 years ago
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