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djyliett [7]
3 years ago
6

△EFG≅△JKL. What is m∠L?

Mathematics
2 answers:
snow_lady [41]3 years ago
4 0
M∠E ≅ m∠J
m∠F ≅ m∠K
m∠G ≅ m∠L

So, m∠ L is 58.
Alex3 years ago
3 0

Answer:

m\angle L=58^{\circ}

Step-by-step explanation:

We have been given an image of two congruent triangles. We are asked to find the measure of angle L.

Since \Delta EFG\cong \Delta JKL, therefore, the corresponding angles will be congruent.

\angle E\cong \angle J

\angle F\cong \angle K

\angle G\cong \angle L

By congruence measure of angle F will be equal to measure of angle G.

Upon looking at our given diagram, we can see that measure of angle G is 58 degrees, therefore, measure of angle L will be 58 degrees as well.

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A company collected $7,800 by selling shoes. The company had $6,084 left after paying the tax. What was the percentage of tax?
dezoksy [38]

Answer:

22%

Step-by-step explanation:

The percentage of tax can be found by calculating the percent change.

new - old / old

6084 - 7800 / 7800

= -0.22 or -22%

This means the price decreased by 22%, meaning you paid 22% in taxes.

please click heart button to give thanks :)

4 0
2 years ago
Show the difference of the two sets
Rudik [331]
There is 8 fruits in set A and 7 in set B
6 0
2 years ago
A rectangular area is to be fenced in with 300 feet of chicken wire. find the maximum area that can be enclosed.
GarryVolchara [31]
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.

2(x+y) = 300
x+y = 150
y = 150-x

A=x(150-x) <--(substitution)

The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150

So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.

A=75(150-75)
A=75*75
A=5625

So the maximum area that can be enclosed is 5625 square feet.
6 0
3 years ago
If &lt; A and &lt; B are vertical angles, and &lt; A = 2x - 11 degrees, and &lt; B = x + 7 degrees, find x.
Tema [17]

Answer:

x=18

Step-by-step explanation:

Vertical angles are congruent. This means that they are equal to each other. So, to solve set them equal to each other and isolate x. First, set up the equation. Then, subtract x from both sides. Finally, add 11 to both sides and simplify.

1) 2x-11=x+7

2) x-11=7

3)x=18

7 0
3 years ago
Un jardín rectangular de 50 cm de largo por 34 m de ancho está rodeado por un camino de arena uniforme.Halla la anchura de dicho
Anastasy [175]

Answer:

5.85 m

Step-by-step explanation:

The width of the sand road can be calculated knowing its area and the dimensions of the rectangular garden as follows:

A_{g} = a.b

<u>Where:</u>

Ag: is the area of the rectangular garden

a: is the length of the rectangular garden = 50 cm = 0.5 m

b: is the width of the rectangular garden = 34 m

A_{s} = 540 m^{2}

<u>Where</u>:

As: is the area of the sand road

The relation between the area of the sand road and the area of the rectangular garden is the following:

A_{s} + A_{g} = (a+2x)*(b+2x)

540 m^{2} + 0.5m*34m = ab + 2ax + 2bx + 4x^{2}

557 m^{2} = ab + 2x(a + b) + 4x^{2}

557 m^{2} - 17m^{2} - 2x(34.5 m) - 4x^{2} = 0

540 - 69x - 4x^{2} = 0                              

By solving the above equation for x we have two solutions:

x₁ = -23.10 m

x₂ = 5.85 m

Taking the positive value, we have that the width of the sand road is 5.85 m.

I hope it helps you!

7 0
3 years ago
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