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Valentin [98]
3 years ago
6

Maths question 11 divided by 4 please give me the answer in decimal as fast as u can

Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0

2.75

Step-by-step explanation:

skad [1K]3 years ago
3 0

\sf\purple{2.75} ✅

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

\frac{11}{4}  \\  = 2.75

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.ඞ}}}}}

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2. Tell whether the segment lengths form an acute, right, or obtuse? 9, 12, 17​
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obtuse angled triangle

Step-by-step explanation:

Given 3 segments , say a, b and c , with c the longest then

• If a² + b² = c² ⇒ right triangle

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• If a² + b² < c² ⇒ obtuse triangle

Here c = 17 and 17² = 289

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Since 225 < 289 , then triangle is obtuse

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What is 90+889+2393+349492+99_495=
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Rowena can paint a room in $14$ hours, while Ruby can paint it in $6$ hours. If Rowena paints for $x$ hours and Ruby paints for
LUCKY_DIMON [66]

Question

Rowena can paint a room in 14 hours, while Ruby can paint it in 6 hours. If Rowena paints for x hours and Ruby paints for y hours, they will finish half of the painting, while if Rowena paints for y hours and Ruby paints for x hours they will paint the whole room. Find the ordered pair (x,y)

Answer:

(x, y) = (231/40, 21/40)

where x = 231/40

y = 21/40

Step-by-step explanation:

From the question, we are told:

Rowena can paint a room in = 14 hours

Ruby can paint it in = 6 hours.

This means

In one hour.

Rowena can paint = 1/14 of the room

Ruby can paint =1/6 hour of the room

From the question, we are told that :

If Rowena paints for x hours and Ruby paints for y hours, they will finish half of the painting

This is represented mathematically as:

(1/14)(x) + ( 1/6)(y) = 1/2...... Equation 1

Also we were told that:

if Rowena paints for y hours and Ruby paints for x hours they will paint the whole room

(1/14)(y) + (1/6)(x) = 1 .......... Equation 2

Bringing the two equations together, we have:

(1/14)(x) + ( 1/6)(y) = 1/2 ....... Equation 1

(1/14)(y) + (1/6)(x) = 1 ............ Equation 2

We find the Lowest common multiple of the numerator 14 and 6 = 42. Hence, we multiply both equations through by 42

3x + 7y = 21 ....... Equation 3

7x + 3y = 42 ......... Equation 4

To solve for x and y in Equation 3 and 4 we would be using the Elimination method

21x + 49y = 147 .......... Equation 5

21x +9y = 126 ........... Equation 6

21x + 49y = 147 .......... Equation 5

-(21x +9y = 126) ........... Equation 6

40y = 21

y = 21/40

To get the value of x , we would substitute 21/40 for y in Equation 3

3x + 7y = 21 ....... Equation 3

3x + 7(21/40) = 21

3x + 147/40 = 21

3x = 21 - 147/40

3x = 693/40

x = (693/40) ÷ 3

x = 693/40 × 1/3

x = 693/120

x = 231/40

Therefore, (x, y) = (231/40, 21/40)

3 0
3 years ago
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