-4y-5>-6+7 with steps please

Mathematics

1 answer:

g100num [7]3 years ago

4 0

Answer:

$y < - \frac{3}{2}$

Step-by-step explanation:

1. Simplify -6+7 to 1.

$- 4y - 5 > 1$

2. Add 5 to both sides.

$- 4y > 1 + 5$

3. Simplify 1+5 to 6.

$- 4y > 6$

4. Divide both sides by -4.

$y < - \frac{6}{4}$

5. Simplify 6/4 to 3/2.

$y < - \frac{3}{2}$

Therefor, the Answer is,

$y < - \frac{3}{2}$

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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})}$