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emmasim [6.3K]
3 years ago
13

Is a triangle a quadrilateral?

Mathematics
2 answers:
Ludmilka [50]3 years ago
6 0
No because it has three sides. a quadrilateral has four sides
Pani-rosa [81]3 years ago
4 0
No quadrilateral has four sides triangle has three
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Thandi received 15% discount on a dress that costs R1500. How much did she<br>pay?​
balandron [24]

Answer:

She paid $1,275.00.

Step-by-step explanation:

I'm assuming you meant $1,500, so I'll be using that.

You can figure this out by subtracting 15% of 1,500 from 1,500.

15% of 1,500 is 225.

1500-225=1275

Therefore, 15% off of $1,500 is $1,275.

5 0
3 years ago
What happens to the force between two charged objects when you triple the magnitude of both charges?
Vika [28.1K]

If the force between two charged objects when you triple the magnitude of both charges. The Force will become 9 times the initial force.

<h3>What is coulombs law?</h3>

According to coulombs law force between two charges is given by  

F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}

Here, R is the distance between both the charges Q and q.

Let the force on the charges be F1

The distance of separation = r

The magnitudes of the charges q1 and q2

K = Coulumb's constant

F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}

Let the force on the charges be F

The distance of separation = r

The magnitudes of the charges 3q1 and 3q2

K = Coulumb's constant

So,

F = \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}

F = \dfrac{1}{4\pi \epsilon }\dfrac{3Q\times 3q}{r^2}\\\\F = \dfrac{1}{4\pi \epsilon }\dfrac{9Qq}{r^2}\\\\F =9  \dfrac{1}{4\pi \epsilon }\dfrac{Qq}{r^2}\\\\F = 9F_1

Learn more about coulombs law;

brainly.com/question/13106909

#SPJ4

8 0
2 years ago
Can someone please help me
rewona [7]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: f(0) = -3

\qquad \tt \rightarrow \: f(2) = 1

\qquad \tt \rightarrow \: f(4) = 1

____________________________________

\large \tt Solution  \: :

F(x) represents the value of y on curve for given value of x

\textsf{\large First -}

\qquad \tt \rightarrow \: f(0) =( 0) {}^{2}  - 3

[ since 0 < 2 ]

\qquad \tt \rightarrow \: f(0) = 0 {}^{}  - 3

\qquad \tt \rightarrow \: f(0) =   - 3

\textsf{\large Second}

\qquad \tt \rightarrow \: f(2) =   (2) {}^{2}  - 3

[ since 2 = 2 ]

\qquad \tt \rightarrow \: f(2) =   4{}^{}  - 3

\qquad \tt \rightarrow \: f(2) =   1

\textsf{\large Third -}

\qquad \tt \rightarrow \: f(4) =    - 4 + 5

[ since 4 > 2 ]

\qquad \tt \rightarrow \: f(4) =    1

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
3x-y=-5<br><br>5x+7y=29<br><br>Solve this for me using substitution pls
pentagon [3]

Answer:

(- 3/13, 56/13)

Step-by-step explanation:

See steps below:)

3 0
3 years ago
(who are you riddle) I'm smaller than 24. I am a common multiple of 7 and 3
sergiy2304 [10]
I think the answer is 21.
6 0
3 years ago
Read 2 more answers
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