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3 years ago

8

Please help I’m really desperate and I will give you a brainliest. And the chooses for number 3 is A .50 , B 1.26, C. 1.41 , D 2

.00

1 answer:

Marizza181 [45]3 years ago

6 0

The answer is D. 40,000÷20,000=2 thus (D). is the answer for problem #3

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a rectangular blue tile has a length of 4.25 inches and a width of 6.75 inches. a similar tile has a length of 12.75 inches.what

Iteru [2.4K]

Answer:

The width of the red tile would be $2.25$ inches.

Step-by-step explanation:

Given blue tile has a length of $4.25$ inches, and a width of $6.75$ inches.

Also, length of similar tile is $12.75$ inches.

Let us assume $x$ as the width of red tile.

It is given in the question that both tiles are similar.

So, their area will be the same.

Area of blue tile would be $4.25\times 6.75=28.6875\ inch^2$

Area of red tile would be $12.75\times x$

The area of blue tile would be equal to area of red tile.

$28.6875=12.75\times x$

$\frac{28.6875}{12.75}=x\\\\x=2.25$

So, the width of the red tile would be $2.25$ inches.

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2 years ago

Find x<br><br> picture listed

never [62]

Answer:

x = 15

Step-by-step explanation:

A full circle = 360°

Therefore,

(8x - 10)° + (6x)° + (10x + 10)° = 360°

Solve for x

8x - 10 + 6x + 10x + 10 = 360

Add like terms

24x = 360

Divide both sides by 24

x = 360/24

x = 15

5 0

2 years ago

What is the slope of a line that is perpendicular to a line whose equation is 3y=−4x+2 ?

vova2212 [387]

we know that

if two lines are perpendicular, then the product of their slopes is equal to minus one

so

$m1*m2=-1$

Step 1

<u>Find the slope of the given line</u>

we have

$3y=-4x+2$

Solve for y

Divide by $3$ both sides

$3y/3=(-4x+2)/3$

$y=-\frac{4}{3}x+ \frac{2}{3}$

the slope of the given line is $m1=-\frac{4}{3}$

Step 2

<u>Find the slope of a line that is perpendicular to the given line</u>

we have

$m1=-\frac{4}{3}$

$m1*m2=-1$

$m2=-1/m1$

substitute the value of m1

$m2=-1/(-4/3)=\frac{3}{4}$

therefore

<u>the answer is</u>

the slope is $\frac{3}{4}$

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3 years ago

Find the Least Common Multiple of these two monomials:

bonufazy [111]

LCM 14 & 24 = 168

then use the highest exponents for the variables

answer = 168x^6y^6z^8

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 <em>equivalence relation</em> 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.

<em>Reflexive: </em>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.

<em>Symmetric</em>: 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.

<em>Transitive</em>: 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 <em>equivalence relation</em>.

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3 years ago