All categories

3 years ago

HELP I NEED SOMEONE TO HELP NO ONE WILL HELP ME

Mathematics

1 answer:

Kobotan [32]3 years ago

3 0

Answer:

Question 1: Answer is H

Question 2: Answer is C

Question 3: Answer is B

Step-by-step explanation:

Trust.

You might be interested in

1/3(22) - 1 -1\2(12)

irinina [24]

The answer is (1/3)...

6 0

3 years ago

Cameron buys 2.45 pounds of apples and 1.65 pounds of pears. Apples and pears each cost c dollars per pound. If the total cost a

nordsb [41]

Answer:

$4.12=2.45c+1.65c$

Step-by-step explanation:

We are given,

Cameron buys 2.45 pounds of apple and 1.65 pounds of pears.

Also, the cost of apples and pears is 'c' dollars per pound.

Thus, the cost of 2.45 pounds of apple is $2.45c$ dollars and the cost of 1.65 pounds of pear is $1.65c$ dollars.

Since, the total cost after using a coupon is $4.12.

So, we get the equation representing the situation is,

Total cost = Total cost of apples + Total cost of pears.

i.e. $4.12=2.45c+1.65c$

i.e. $4.12=4.1c$

i.e. $c=\frac{4.12}{4.1}$

i.e. c = 1 dollar

Hence, the required equation to find c is $4.12=2.45c+1.65c$ .

5 0

3 years ago

I really need help with this ASAP

Alexandra [31]

Answer:

No. of boxes = 10

Oranges per box = 56

Total no. of oranges = 56*10=560

No. of bad oranges = 560/40= 14

(i) Probability of bad orange = 14/560 = 1/40

(ii) no of oranges expected to be bad = 14 ( found above)

Hope it helps.............. :)

7 0

3 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get