Ask question

Ask question

All categories

3 years ago

6

Tartan and Alastair both mow lawns each charges a flat fee to mow a lawn Table number of past time spent on money

1 answer:

sineoko [7]3 years ago

8 0

What's the question?

You might be interested in

Please Evaluate 27 times ( 1/3) to the 3 power. A). 1 B). 3 C). 9 D). 27

topjm [15]

Answer:

you want to follow PEMDAS so you would multiply 27 by 1/3 to get 81.003, which you would round to 81, then you would multiply 8 to the third power and you would get 512.

Step-by-step explanation:

27(1/3)^3

81^3

512

7 0

3 years ago

What is the difference between 10 pounds and 10 kilograms

katen-ka-za [31]

Both pound and kilogram are units of measurement of weight or mass. A pound is an imperial unit of mass or weight

7 0

2 years ago

Read 2 more answers

at 4:00 am the temperature was -9 degrees f. at noon the temperature was 18 degrees f. what was the change in temperature

Nutka1998 [239]

There was a change of 27 degrees F.

7 0

3 years ago

Help please i need to know this

Harlamova29_29 [7]

Answer:

As shown in the attachment,

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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

2 years ago