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Tomtit [17]
2 years ago
8

Given: ABCDis a parallelogram ∠GEC ≅ ∠HFA and AE ≅FC. Prove △GEC ≅ △HFA.

Mathematics
2 answers:
alexandr402 [8]2 years ago
7 0

See the proof in the image

OlgaM077 [116]2 years ago
4 0

<G E C=<H F A

A s

  • A E=F C
  • A F=E C

A lso

  • <CAH=<GCE

Hence

  • △GEC ≅ △HFA(ASA)
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Eugene walked all the way to school at 3 mph then realized she forgot her math book (how could she?!), so she ran back at 7 mph.
Fynjy0 [20]

\text{Answer: }1\frac{23}{40}\ miles\text{ far she live from school.}

Explanation:

Since we have given that

Speed at which Eugene walked all the way to school = 3 mph

After realizing that she forgot her math book, she ran back

Speed at which Eugene walked back to her house = 7 mph

Let the distance between her house and her school be x

According to question,

\frac{x}{3}+\frac{x}{7}=\frac{45}{60}\\\\\frac{7x+3x}{21}=\frac{3}{4}\\\\\frac{10x}{21}=\frac{3}{4}\\\\x=\frac{3\times 63}{4\times 10}=\frac{63}{40}=1\frac{23}{40}\ miles.

Hence,

1\frac{23}{40}\ miles\text{ far she live from school.}

8 0
3 years ago
Chris can type 490 words in 7 min how many works can he type PER min <br> You get 14 points
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<h2>Answer: 70 words per min</h2><h2 /><h2>Step-by-step explanation: We just divided 490 by 7 to get how many words he can type for 1 min!</h2>

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3 years ago
If f(x) = 3^x+10x and g(x) = 4x-2, find (f-g)(x)
DENIUS [597]
(f-g)(x) = f(x) - g(x) = 3^x + 10x - (4x - 2)  = 3^x + 6x + 2 Answer
4 0
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Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”
Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± \sqrt{1-(y-2)^{2}}

y = ± \sqrt{1-x^{2}}+2

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

 R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

 of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

  (x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒  (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± \sqrt{1-(y-2)^{2}}

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± \sqrt{1-x^{2}}

- Add 2 for both sides

∴ y = ± \sqrt{1-x^{2}}+2

∴ y = h(x)

- In the function x = ± \sqrt{1-(y-2)^{2}}

∵ \sqrt{1-(y-2)^{2}} ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

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∵ \sqrt{1-x^{2}} ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0
3 years ago
A company has 875 employees. "Half - Price Wednesday," 64% of the employees eat lunch at the company cafeteria. How many employe
Vika [28.1K]
875 × .64= 560

answer: 560 employees eat lunch at the cafeteria on Wednesday.
4 0
3 years ago
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