Brian pays £475.29 a year on his car insurance. The insurance company reduces the price by 2.1%.

We apply the slope formula by substituting the given points.

Here, x2 equals 5 and x1 equals -2; y2 equals -3 and y1 equals 6.

Thus, the line's slope is -9 divided by 7.

Result:

6.1°; 425.86 m.

Step-by-step breakdown:

The information provided states that the airplane is at an altitude of 5.7 miles above ground level, while the "radius of Earth is about 4000 miles." Thus,

θ = 2 × cos^-1 (a/ (a + b)), where a = 4000 miles, and b = 5.7 miles.

θ = 2 × cos^-1 (4000/ (4000 + 5.7)) = 6.1°.

To calculate the distance in meters:

Change in distance = 6.1° /360° × 2π × 4000 miles = 425.86 meters.

Consequently, BD⌢ measures at 6.1° and the distance corresponding to this section of Earth is 425.86 meters.

Answer:

The blimp's altitude is 1635.363039 feet above the ground.

Step-by-step explanation:

Archie and Veronica stand one mile (5280 ft) apart on the ground, leading to AV = 5280 feet.

Let's denote the blimp's height above ground as XB = h feet.

Archie records the angle of elevation from the ground to the blimp as 53 degrees, signifying angle ∠BAX = 53°.

Veronica, meanwhile, measures the angle of elevation to the blimp at 22 degrees, indicating angle ∠BVX = 22°.

In Right triangle ΔAXB, we utilize cot(A) = AX / XB.

Thus, AX can be expressed as AX = XB*cot(A) = h*cot(53°) = 0.75355405*h.

In Right triangle ΔVXB, cot(B) is similar: cot(B) = VX / XB.

Consequently, VX is VX = XB*cot(B) = h*cot(22°) = 2.475086853*h.

We know AV = 5280, meaning AX + VX = AV.

Therefore, AX + VX equals 5280.

From here, we can write: 0.75355405*h + 2.475086853*h = 5280.

This simplifies to: 3.228640904*h = 5280.

Thus, h is calculated as 5280 / 3.228640904 = 1635.363039 feet.

Thus, the blimp's height from the ground is 1635.363039 feet.

Answer:

We will mix 2.4 mL of water with 0.6 mL of the coloring solution.

The total solution volume amounts to 2.4 + 0.6, which equals 3 mL.

Therefore,

0.6 mL of coloring solution now contains 2.4 mL of water.

Alternatively,

1 mL of this coloring solution comprises 4 mL of water.

Thus,

We should prepare 3 mL of the coloring solution with the resultant water.

This 3 mL mixture will encapsulate 0.6 mL of the coloring solution.

Step-by-step explanation:

As stated:

The formula requires 0.6 mL of a coloring solution.

Using a 10-mL graduated cylinder that is divided from 2 to 10 mL in 1-mL increments:

We aim to apply the aliquot method to accurately measure 0.6 mL of the coloring solution.

Given that the graduated cylinder measures from 2mL to 10mL only, we'll blend 2.4 mL of water with 0.6 mL of coloring solution.

Thus, the total solution volume is 2.4 + 0.6, resulting in 3 mL.

Consequently,

This implies that 0.6 mL of coloring solution now carries 2.4 mL of water.

Additionally,

1 mL of coloring solution encapsulates 4 mL of water.

Hence,

We will quantify 3 mL of the final solution, which contains 0.6 mL of coloring solution.