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Misha Larkins [42]
2 years ago
5

What is the domain of the following parabola? (5 points)

Mathematics
1 answer:
nikitadnepr [17]2 years ago
6 0

Answer:

the domain for this parabola is all real numbers

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A rectangle with an area of 4/7 m2 is dilated by a factor of 7. What is the area of the dilated rectangle
tatyana61 [14]

according to the picture we have:

x.y=4.7

(7)x(7)y=49xy=(49)(4.7)=230.3

6 0
3 years ago
59:26 What is the length of the hypotenuse, x, if (20, 21, x) is a Pythagorean triple? 22 29 41 42
Vikki [24]
If the legs are length a and b and hyptonuse is c then
a²+b²=c²

so
if the legs are 20 and 21 and the hypotnuse is x then
20²+21²=x²
400+441=x²
841=x²
29=x
4 0
3 years ago
Read 2 more answers
Need help badly don’t have answer to this problem.
NikAS [45]
There are a couple of ways to tackle this one, using the 45-45-90 rule or just using the pythagorean theore, let's use the pythagorean theorem.

the angle at A is 45°, and its opposite side is BC, the angle at C is 45° as well, and its opposite side is AB, well, the angles are the same, thus BC = AB.

hmmm le'ts call hmmm ohh hmmm say z, thus BC = AB = z.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2 \implies (6\sqrt{2})^2=z^2+z^2
\qquad 
\begin{cases}
c=\stackrel{6\sqrt{2}}{hypotenuse}\\
a=\stackrel{z}{adjacent}\\
b=\stackrel{z}{opposite}\\
\end{cases}\\\\\\ 6^2(\sqrt{2})^2=2z^2\implies 36(2)=2z^2\implies \cfrac{36(2)}{2}=z^2\implies 36=z^2
\\\\\\
\sqrt{36}=z\implies 6=z
4 0
3 years ago
Read 2 more answers
*In ∆ABC, on the extension of the side BC , draw a line segment CD ≅ CA . Draw the segment AD . The line segment CE is the angle
Leya [2.2K]

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Step-by-step explanation:

In ∆ABC

  • On the extension of the side BC , draw a line segment CD ≅ CA
  • Draw the segment AD
  • The line segment CE is the angle bisector of ∠ACB
  • The line segment CF is the median towards AD in ∆ ACD

We want to prove that CF ⊥ CE

Look to the attached figure

In Δ ABC

∵ CE is the bisector of angle ACB

∴ ∠ACE ≅ ∠BCE

In Δ ACD

∵ CA = CD

∴ Δ ACD is an isosceles triangle

∵ AD is the median towards AD

- In any isosceles triangle the median from a vertex to its opposite

  side bisects this vertex

∴ AD bisects ∠ACD

∴ ∠ACF ≅ ∠DCF

∵ BCD is a straight segment

∵ CE , CA , CF are drawn from point C

∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°

∵ m∠ACE ≅ m∠BCE

∵ m∠ACF ≅ m∠DCF

- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF

∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°

∴ 2 m∠ACE + 2 m∠ACF = 180°

- Divide all terms by 2

∴ m∠ACE + m∠ACF = 90°

∴ EC ⊥ CF

CE ⊥ CF because m∠ECA + m∠FCA = 90°

Learn more:

You can learn more about perpendicular lines in brainly.com/question/11223427

#LearnwithBrainly

7 0
3 years ago
An airplane travels 6903 kilometers against the wind in 9 hours and 8613 kilometers with the wind in the same amount of time. Wh
Rzqust [24]

Answer:

The rate of the plane is 862 miles per hour

Step-by-step explanation:

An airplane travels 6903 kilometers against the wind in 9 hours.

Speed = distance / time

Speed of the airplane while travelling against the wind is

6903/9 = 767 miles per hour

The airplane travelled 8613 kilometers with the wind in the same amount of time. This means that the speed while travelling with the wind will be

8613/9 = 957 miles per hour

Let x = speed of the airplane

Let y = speed of the wind

While travelling against the wind,

x - y = 767 - - - - - -- 1

While travelling with the wind,

x + y = 957 - - - - - - -2

Subtracting equation 2 from equation 1,

-y - y = 767 - 957

-2y = -190

y = -190/-2

y = 95 miles per hour

x = 957 - y

x = 957 - 95

x = 862 miles per hour

4 0
3 years ago
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