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4 years ago

12

ALGEBRA 1, SEM. 2

1 answer:

grigory [225]4 years ago

3 0

2•(1/4)^(2-1)= 2•(1/4)= 1/2
2•(1/4)^(3-1)= 2• (1/4)^2 = 2•1/16=1/8

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Whats 1+1<br> your welcome

Vaselesa [24]

Answer:

2 :)

Step-by-step explanation:

let's say you have 1 apple, if you add another apple to the pile, you will have 2 apples

if you remove the apples at the end, you get the equation 1 + 1, and now you know 1 + 1 = 2

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2 years ago

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forty children from an after school club went to the matinee. this is 25% of the children at the club. how many children are in

Aleks [24]

Number of children from an after school that went to the martinee = 40
Percentage of children from an after school that went to the martinee = 25%
Let us assume the number of children in the club = x
Then
25% * x = 40
(25x/100) = 40
x/4 = 40
x = 40 * 4
= 160
So the total number of children in the club is 160. I hope the procedure is clear enough for you to understand. you can always use this method for solving similar problems without requiring any help from outside.

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3 years ago

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Guys please help me with this?!

ser-zykov [4K]

Answer:

X = 41

y = 7.22

Step-by-step explanation:

5 0

3 years ago

Simplify 9+(-2) to the third power<br><br> 1<br> 3<br> 17

AleksandrR [38]

Answer:

1

Step-by-step explanation:

9+ (-2)^3

We need to do exponents first according to PEMDAS

9 + (-8)

Then we add

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7 0

3 years ago

Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i,2 +2i,5i; (b)2i,3

Illusion [34]

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

Use the distance formula to find the distance between two points

$AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}$

$BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}$

$CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}$

For the triangle to be right angles triangle

$BC^{2}=AB^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

$AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}$

$BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}$

$CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}$

For the triangle to be right angles triangle

$AB^{2}=BC^{2}+CA^{2}$

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

$AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}$

$BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}$

$CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}$

For the triangle to be right angles triangle

$CA^{2}=BC^{2}+AB^{2}$

Here, it is valid, so these are the points of a right angled triangle.

7 0

3 years ago