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3 years ago

BRAINLIEST!!!! to best answer

Mathematics

1 answer:

ad-work [718]3 years ago

6 0

B. This is clear because of using the intercepts method. You know it crosses at y = 3 and x = -7. You also know to shade over by testing a spot in the shaded area.

You might be interested in

Tell whether each number is rational or irrational: Pi, -1/2, the square root of 2/3, and 0.57575757

DedPeter [7]

Pi: Irrational
√(2/3): Irrational

0.57575757 is rational
and -1/2 is rational.
I hope this helps.

5 0

3 years ago

Against the wind a small plane flew 715 miles in 2 hours and 10 minutes. The return trip took only 1 hour and 50 minutes. What w

ZanzabumX [31]

Answer:

360 mi/h

Step-by-step explanation:

The speed for the outbound trip was ...

speed = distance/time = (715 mi)/(2 1/6 h) = 330 mi/h

The inbound speed was ...

(715 mi)/(1 5/6 h) = 390 mi/h

The airspeed of the plane is the average of these two ground speeds, so is ...

(330 +390)/2 = 360 . . . . mi/h

8 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

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

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

3 years ago

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w

Liono4ka [1.6K]

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds. Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

$t = 5.0625$