Solve the following systems of inequalities: {-5x+3y≤45 {2x+3y>24

Sabrina has a rectangular swimming pool in her backyard. She fills it with water to the depth of 4 feet. The water in the pool h

aleksklad [387]

Answer: $20\times 20, 16\times 25, 10\times 40$

Step-by-step explanation:

Given

The volume of the pool is $1600\ ft^3$

Depth of water is $d=4\ ft$

Suppose A is the base of the pool

$\Rightarrow V=A\times d\\\Rightarrow 1600=A\times 4\\\Rightarrow A=400\ ft^2$

This area can be written as

$\Rightarrow A=10\ ft\times 40\ ft\ or\\\Rightarrow A=16\ ft\times 25\ ft\ or\\\Rightarrow A=20\ ft\times 20\ ft$

7 0

3 years ago

Solve the equation and this is an assignment btw

lilavasa [31]

Answer:

z=6 and z=-6

Step-by-step explanation:

$7z^2=252$

Divide both sides by 7

$\frac{7z^2}{7} =\frac{252}{7} \\z^2=36$

Find the square root

z=±√36

z=±6

Therefore, z=6 and z=-6.

I hope this helps!

4 0

3 years ago

If you take a number, times by 8 then add 6. You get the same as if you took the number multiply by 3 then subtract 7. What is t

igor_vitrenko [27]

I’m not sure this question is kinda confusing.

8 0

3 years ago

Anyone Can help me? Thanks

Elena L [17]

Answer:

9.8

Step-by-step explanation:

updated

9^2=x^2+4^2

9*9=x*x+4*4

81=x*x-16

+16. +16

97=x*x

√97=√x*x

√97=x

So the answer is √97, but the question wants it rounded so it's actually 9.8

5 0

4 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

$A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)$ .

Recall that if P and Q are invertible, then QP is invertible and $(QP)^{-1}=P^{-1}Q^{-1}$ . So, if we denote R=QP we obtained that

$A=R^{-1}CR$ . Hence, A↔C.

Therefore, the relation is an equivalence relation.

4 0

4 years ago