Ask question

Ask question

All categories

2 years ago

13

Sheldon wants to access the money in this checking account. Which of the following will work?

1 answer:

bulgar [2K]2 years ago

7 0

D. he can access it by taking it out of his account

You might be interested in

At a wedding, there are 456 people spread out amongst 45 tables. There are no empty seats. The reception hall has tables that si

uysha [10]

Thats a lot of people for one resturant

3 0

3 years ago

PLEASE HELP ILL DO ANYTHING

yuradex [85]

Answer:

B) 2 1/4

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Round to the nearest hundredth 5/8

Solnce55 [7]

Eighths and their multiples are common fractions which I recommend memorizing, but to actually solve this, you use the literal meaning of a fraction and divide 5 by 8. See the long-division below (it was surprisingly difficult to type, so I hope it helps!).

To round 0.625 to the nearest hundredth, we go to the second decimal place, which is 5, so we round up to 0.63.

3 0

2 years ago

ALGUEMMM ME AJUDAAA POR FAVORRR!!!!!!!!!!!!!!!!!!!

Amanda [17]

Answer:

50)

a. + 13

B. +10

C. -15

D. -5

Step-by-step explanation:

8 0

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

2 years ago