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3 years ago

What is the scale factor from abc to def?

Mathematics

1 answer:

vlabodo [156]3 years ago

3 0

Answer:

1/3

Step-by-step explanation:

Considering the triangles are similar the side lengths would be proportional too

like AB is similar to DE so 5/15 = 1/3

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Please help if so thank you And explain

Naddika [18.5K]

Answer:

x = 34°

Step-by-step explanation:

Given AC and BD are perpendicular bisectors, we can say that at point E, there are 4 right angles [perpendicular bisectors intersect to create 4 90 degree angles].

Now, if we look at the triangle AED, we know that it is a right triangle, meaning that angle E is a right angle.

Also,

We know sum of 3 angles in a triangle is 180 degrees. Thus, we can write:

∠A + ∠E + ∠D = 180

<em>Note: Angle A and Angle D are just the half part of the diagram. More exactly we can write:</em>

∠EAD + ∠ADE + ∠DEA = 180

Given,

∠EAD = 56

∠DEA = 90

We now solve:

∠EAD + ∠ADE + ∠DEA = 180

56 + ∠ADE + 90 = 180

146 + ∠ADE = 180

146 + x = 180

x = 180 - 146

x = 34°

4 0

3 years ago

4/10 + 3/16 = what is the answer of the addition fraction

natka813 [3]

Answer:

4/10+3/16=47/80

Step-by-step explanation:

(4/10×8/8)+(3/16×5/5)=?

32/80+15/80=?

32+15/80=47/80

3 0

2 years ago

Solve the equation by completing the square. ^x2- 12x = -28

Olegator [25]

x^2 -12 x = -28

(-12/2) ^ 2 =36

x^2 -12 x + 36 = -28 + 36

x^2 -12x + 36 = 8

(x-6)^2 = 8

take the square root of each side

x-6 = sqrt (8) x-6 = -sqrt (8)

you get a positive and a negative when taking the square root

x= 6 + sqrt (8)

x = 6 - sqrt (8)

sqrt (8) = 2 sqrt (2)

x= 6 + 2 sqrt (2)

x = 6 - 2 sqrt (2)

Answer:

x= 6 + 2 sqrt (2)

x = 6 - 2 sqrt (2)

3 0

2 years ago

Read 2 more answers

Peter's farm has 160 meters of fencing, and he wants to fence a rectangular field thatborders a straight river. He needs no fenc

Likurg_2 [28]

ANSWER

3200 m²

EXPLANATION

Peter wants to build a fence around a rectangular field, except for one side because the river is there,

So, the total perimeter of the fence is,

$P=W+W+L=2W+L$

And this is equal to 160 m. Solving for L,

$160=2W+L\text{ }\Rightarrow\text{ }L=160-2W$

The area of this field is,

$A=WL$

Replace L with the expression we found from the perimeter,

$A=W(160-2W)=160W-2W^2$

The area is given by a quadratic function whose leading coefficient is negative, which means that the graph is a downward parabola and, therefore, the vertex is the maximum value of the area.

The x-coordinate of the vertex is given by,