More than half of which climate type averages between 40 and 59 inches of precipitation a year?

Kiko is coasting on her bicycle down a hill. After 3 s, her speed is 10 m/s. After 8 s, her speed is 25 m/s. Her acceleration during this time will be  5 m/s².

Acceleration is the rate of change in both speed and direction of velocity over time. When anything moves faster or slower in a straight line, it is said to have been accelerated. Because the direction is always shifting, motion on a circle accelerates even while the speed is constant.

It is given that, Kiko is coasting on her bicycle down a hill. After 3 s, her speed is 10 m/s. After 8 s, her speed is 25 m/s.

Acceleration is found as the ratio of the change in speed to the change in a time interval.

Acceleration =  change in speed/ change in a time interval

Acceleration = (25-10) / (8-5)

Acceleration = (15)/(3)

Acceleration = 5 m/s²

Thus, Kiko is coasting on her bicycle down a hill. After 3 s, her speed is 10 m/s. After 8 s, her speed is 25 m/s. Her acceleration during this time will be  5 m/s².

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Division Property of Equality

16m = 160

 divide both sides by 16 to find m

m = 160/16 = 10

Answer:

Step-by-step explanation:

The given expression is

[tex]16m=160[/tex]

So, we use one of the properties of equalities for division, which states that if we divide one side of the equation, we also must divide the other side by the same number, to not mess up the balance of the equation. Applying this property, we have

[tex]\frac{16m}{16}=\frac{160}{16}\\ m=10[/tex]

The question revolves around solving a system of three linear equations with three variables, using algebraic techniques such as substitution or elimination.

The system of equations provided can be solved using methods such as substitution, elimination, or matrix operations. The equations given are:

To solve this system, we can perform operations such as eliminating one variable at a time or using matrices to find the values of x, y, and z that satisfy all three equations simultaneously.

Since Ned's truck suffered body damage due to a hailstorm, the components of Ned's insurance policy will take care of the damage done to Ned's truck during the hailstorm is comprehensive damage.

A comprehensive coverage is an insurance policy which covers the cost of damages to a vehicle which has been involved in an accident that is not caused by a collision.

Comprehensive insurance covers losses such as:

Therefore, the components of Ned's insurance policy will take care of the damage done to Ned's truck during the hailstorm is comprehensive damage.

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we have

[tex]f(x)=(x-3)^{2}(x+2)^{2}(x-1)[/tex]

we know that

The zero of the function is  the value of x when the value of the function is equal to zero

[tex]f(x)=(x-3)^{2}(x+2)^{2}(x-1)[/tex]

rewrite the function

[tex]f(x)=(x-3)(x-3)(x+2)(x+2)(x-1)[/tex]

The zero's of the function are

[tex]x=3\ x=3\ x=-2\ x=-2\ x=1[/tex]

so

[tex]x=3[/tex] ------> has a multiplicity of [tex]2[/tex]

[tex]x=-2[/tex] ------> has a multiplicity of [tex]2[/tex]

[tex]x=1[/tex] ------> has a multiplicity of [tex]1[/tex]

therefore

the answer is

The zero [tex]1[/tex] has a multiplicity of [tex]1[/tex]

The zero [tex]-2[/tex] has a multiplicity of [tex]2[/tex]

From the given function, f(x) = (x − 3)2(x + 2)2(x − 1);

The function f(x) = (x − 3)²(x + 2)²(x − 1) given can be expanded so that we have;

Hence, the zeros of the function f(x) are;

We can conclude that the zero 1 has a multiplicity of 1 and the zero -2 has a multiplicity of 2.

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To earn $500 in interest at a 4% annual rate, you need to leave $2000 in the account for approximately 6.25 years.

The formula for simple interest is given by:

[tex]\[ I = P \times r \times t \][/tex]

Where:

- I is the interest earned,

- P is the principal amount (initial deposit),

- r is the annual interest rate (expressed as a decimal), and

- t is the time the money is invested or borrowed in years.

In this case, you want to find t, so rearrange the formula to solve for t:

[tex]\[ t = \frac{I}{P \times r} \][/tex]

Given that P = $2000, r = 0.04 (4% as a decimal), and I = $500, plug these values into the formula:

[tex]\[ t = \frac{500}{2000 \times 0.04} \][/tex]

Calculate the result to find out how long you must leave the money in the account to earn $500 in interest.

[tex]\[ t = \frac{500}{2000 \times 0.04} \]\[ t = \frac{500}{80} \]\[ t = 6.25 \][/tex]

Therefore, you must leave the money in the account for approximately 6.25 years to earn $500 in interest at a simple annual rate of 4%.

Answer:

The remainder is 80

Step-by-step explanation:

Given the polynomial

[tex]f(x)=4x^3-5x^2+2x+11[/tex]

we have to find the remainder when the above polynomial f(x) is divided by x-3

By remainder theorem when we put x=3 in f(x) we get the remainder i.e f(3)

[tex]f(3)=4(3)^3-5(3)^2+2(3)+11[/tex]

[tex]=108-45+6+11[/tex]

[tex]=80[/tex]

which is required remainder.

Hence, the remainder is 80

The root of the function is such point [tex]x_0,[/tex] for which [tex]f(x_0)=0.[/tex]

The attached graph of the function  [tex]f(x) = 2x^3 - 19x^2 + 57x - 54[/tex] shows, that the graph of the function intersects the x-axis in three points (2,0), (3,0) and (4.5,0).

This means that for

Thus, x=2, x=3 and x=4.5 are rational roots of the polynomial function f(x).

Answer: three rational roots.

The polynomial function f(x) = 2x^3 - 19x^2 + 57x - 54 has three rational roots at x=2, x=3, and x=4.5, as confirmed by the graph intersecting the x-axis at these points.

The graph of the function f(x) = 2x^3 - 19x^2 + 57x - 54 intersects the x-axis at three points: (2,0), (3,0), and (4.5,0). This implies that x=2, x=3, and x=4.5 are the roots of the polynomial function.

Since these values are rational, the function has three rational roots. Rational roots occur when the function evaluates to zero for specific rational input values.

In this case, f(2) = 0, f(3) = 0, and f(4.5) = 0. The graph crossing the x-axis at these points confirms their status as roots. The polynomial f(x) has three rational roots, which are x = 2, x = 3, and x = 4.5.

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t = time

v = velocity

you have (2,54) & (4,58)

v-54 = (58-54) / (4-2) *(t-2)

v=4/2(t-2)+54

v =2(t-2)+54

v = 2t-4+54

b - Graph by plotting the given points (2,54) & 4,58) and drawing a straight line through them