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3 years ago

Need help asap ! thanks

Mathematics

2 answers:

Softa [21]3 years ago

5 0

3.33 is the right answer

poizon [28]3 years ago

4 0

6 is the correct answer you have to combine like terms and take 16 out of the equation which leaves you with -6n=36 so if you change n to 6 it is 36

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Is the square of an irrational number always irrational? Explain

iVinArrow [24]

<h2>Answer:</h2><h2><u>NOO</u></h2>

Step-by-step explanation:

<h3>For example, √3 is irrational, but (√3)2 = 3, which is rational.</h3>

HOPE IT HELP

7 0

3 years ago

Read 2 more answers

Help needed now with Math question.

saveliy_v [14]

Answer:

is the last one

Step-by-step explanation:

5 0

3 years ago

-8 = -8 (5-2x) please explain

Lerok [7]

Answer:

x= 2

Step-by-step explanation:

-8(5)(-2x)

-40 16x

+40

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5 0

3 years ago

A space plane skims the edge of space at 4000 miles per hour. Neglecting altitude, if the circumference of the planet is approx

posledela

Answer:

6 hour 20 minutes

Step-by-step explanation:

We are given that a space plane skims the edge of space at 400 miles per hour.

Circumference of the planet=25000 miles approximately

We have to find the time taken by the plane to travel around the planet.

Speed of plane=4000 miles per hour

Total distance traveled by the plane= approximately 25000 miles

Because plane travel in circular path therefore, distance

traveled by plane=Circumference of planet

Time = $\frac{distance}{speed}$

Substitute the values then we get

Time = $\frac{25000}{4000}=\frac{25}{4}=6\frac{1}{4} hours$

Time= $6 hour +\frac{1}{4}\times 60 minutes= 6\;hour\;20 minutes$

1 hour=60 minutes

Hence, the plane takes time to travel around the planet=6 hour 20 minutes

7 0

3 years ago

Three bouquets of flowers are ordered at a florist. 3 roses, 2 carnations, and 1 tulip cost $14. 6 roses, 2 carnations, and 6 tu

STALIN [3.7K]

cost of 1 rose = $ 3

cost of 1 carnation = $ 1

cost of 1 tulip = $ 3

<em><u>Solution:</u></em>

Let "r" be the cost of 1 rose

Let "c" be the cost of 1 carnation

Let "t" be the cost of 1 tulip

<em><u>3 roses, 2 carnations, and 1 tulip cost $14</u></em>

So we can frame a equation as:

3 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 14

$3 \times r + 2 \times c + 1 \times t = 14$

3r + 2c + 1t = 14 ----- eqn 1

<em><u>6 roses, 2 carnations, and 6 tulips cost $38</u></em>

So we can frame a equation as:

6 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 6 tulip x cost of 1 tulip = $ 38

$6 \times r + 2 \times c + 6 \times t = 38$

6r + 2c + 6t = 38 ------ eqn 2

<u><em>1 rose, 12 carnations, and 1 tulip cost $18</em></u>

So we can frame a equation as:

1 rose x cost of 1 rose + 12 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 18

$1 \times r + 12 \times c + 1 \times t = 18$

r + 12c + t = 18 ----- eqn 3

<em><u>Let us solve eqn 1 and eqn 2 and eqn 3 to find values of "r" "c" "t"</u></em>

3r + 2c + 1t = 14 ----- eqn 1

6r + 2c + 6t = 38 ------ eqn 2

r + 12c + t = 18 ----- eqn 3

From eqn 1,

3r = 14 - 2c - t

$r = \frac{14 - 2c - t}{3}$

Substitute the above value of r in eqn 2

$6(\frac{14 - 2c - t}{3})+ 2c + 6t = 38\\\\28 - 4c - 2t + 2c + 6t = 38\\\\-2c +4t = 10\\\\-2c = 10 - 4t\\\\2c = 4t - 10\\\\c = 2t - 5$

Substitute c = 2t - 5 and $r = \frac{14 - 2c - t}{3}$ in eqn 3

$12(2t - 5) + \frac{14 - 2c - t}{3} + t = 18\\\\24t - 60 + \frac{14-2(2t - 5) - t}{3} + t = 18\\\\72t - 180 + 14 - 4t +10 - t + 3t = 54\\\\70t = 54 + 180 - 14 -10\\\\70t = 210\\\\t = 3$

Substitute t = 3 in c = 2t - 5

c = 2(3) - 5

Substitute t = 3 and c = 1 in $r = \frac{14 - 2c - t}{3}$

$r = \frac{14 - 2(1) - 3}{3}\\\\r = \frac{14 - 2 - 3}{3}\\\\r = \frac{9}{3} = 3$

<em><u>Summarizing the results:</u></em>

cost of 1 rose = $ 3

cost of 1 carnation = $ 1

cost of 1 tulip = $ 3

6 0

3 years ago