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Juli2301 [7.4K]

2 years ago

11

A train travels in a straight path at a steady speed of 1.25 miles per minute. at noon, the train is 100 miles from the starting

point. express the distance d from the starting point as a function of time, x, in minutes after noon. d = 100 2.5 x d = 1.25 x d = 1.25 100 x d = 100 1.25 x

1 answer:

gavmur [86]2 years ago

8 0

Correct option is D. Distance d = 100 + 1.25 x

<h3>What is a linear equation?</h3>

A linear equation only has one or two variables. No variable in a linear equation is raised to a power greater than 1 or used as the denominator of a fraction. When you find pairs of values that make a linear equation true and plot those pairs on a coordinate grid, all of the points lie on the same line.

An example of linear equation is y=mx + b. A polynomial equation of the first degree (such as x = 2y - 7)
A linear equation is an algebraic equation where each term has an exponent of 1 and when this equation is graphed, it always results in a straight line. This is the reason why it is named as a 'linear equation'.

To lean more about linear equations from the given link

brainly.com/question/2030026

#SPJ4

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Find the value of a square + b square when a+b = 6 a_b = 4

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3 years ago

1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

8 0

3 years ago