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3 years ago

ANSWER QUICKLY PLEASE. WILL BE AWARDED BRAINLIEST ANSWER

Mathematics

1 answer:

Alexxandr [17]3 years ago

8 0

Area is found by multiplying the length by the width.

Let X = width.

The length is x +3 ( 3 cm longer than the width).

Now you have area = x * x+3

Area is given as 754 square cm.

754 = x * x+3

Simplify:

754 = x^2 +3x

Subtract 754 from both sides:

x^2 +3x - 754 = 0

Factor:

(x -26) (x+29)

Solve for x:

x = 26 and -29.

Since the dimension cannot be negative, x = 26.

The width = 26 cm.

The length = 26 +3 = 29 cm.

Multiply each dimension by 100 to get millimeters:

Width = 260 mm

Length = 290 mm

You might be interested in

If y varies directly as x, and y is 6 when x is 72, what is the value of y when x is 8?

Eddi Din [679]

Answer:

2/3

Step-by-step explanation:

Y varies directly as x

Y =kx

Y=6 when x = 72

First find the relationship between y and x.

i.e find the value of k

Substitute the value of x and y

6 = k *72

Make k the subject of the formula

K = 6/72

K= 1/12

:. The rlationship between x and y is

Y = 1/12x

When x is 8

Y = 1/12 * 8

Y = 8/12

Y = 2/3 and that is the answer

Hope it helped you

3 0

3 years ago

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Tia’s car needs repairs. Honest Harry will charge $70 per hour plus $130 for the part. Lucky Lou will charge $80 per hour plus $

elena-s [515]

To answer the question above, let x the number of hours for the two jobs to cost equal. Harry's charge can be expressed as 70x + 130 and that of Lou is 80x + 40. Equating both,

70x + 130 = 80x + 40 ; x = 9

Thus, the job should take 9 hours for both charges to be equal.

3 0

3 years ago

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An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the he

timama [110]

Answer:

2330.51 miles

Step-by-step explanation:

Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees, $d_1 = 1\times610=610$ miles.

The distance covered at 310 degrees, $d_2 = 3\times610=1830$ miles.

Total distance covered, d, is the magnitude of the sum of vectors $\vec{d_1}$ and $\vec{d_2}$ as shown in the figure.

The angle between the vectors $\vec{d_1}$ and $\vec{d_2}$ , $\theta=310-270=40$ degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

$d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^{\circ})}$