1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Jet001 [13]
2 years ago
9

Figure A was drawn first and figure B was drawn second. What is the scale factor?

Mathematics
1 answer:
kakasveta [241]2 years ago
3 0
The answer is A because you are multiplying it by 2
You might be interested in
Ellie wanted to build a fence around her rectangular garden. The garden is 10 ft by 20 ft. A) What is the perimeter of the garde
swat32

Answer:

60 feet

Step-by-step explanation:

Perimeter is the sum of the four sides of a rectangle

the formula for the perimeter of a rectangle = 2 x (length + breadth)

2 x (10 + 20)

2 x (30)

= 60 feet

Alternatively, you can add the sides together : 10 + 10 + 20 + 20 = 60 feet

4 0
2 years ago
Solve for Variable: 3x + 7 &lt; 16<br> A. x &gt; 3<br> B. x &lt; 5<br> C. x &gt; 6<br> D. x &lt; 3
romanna [79]

Answer:

D. x < 3

Step-by-step explanation:

3x + 7 < 16

To solve, we have to isolate x.

3x + 7 < 16

Subtract 7 from both sides :

3x < 9

Divide 3 from both sides :

x < 3

Now in some cases, you could say that the inequality sign would switch, but this is only when you divide by a negative sign, so keep that in mind.

4 0
2 years ago
How to do the inverse of a 3x3 matrix gaussian elimination.
nata0808 [166]

As an example, let's invert the matrix

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}

We construct the augmented matrix,

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]

• Multiply row 3 by 1/3 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 2 by 1/7 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 1 by -1/3 :

\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

So, the inverse of our matrix is

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}

6 0
2 years ago
Explain the purpose of statements and reasons in a formal proof.
baherus [9]

Geometric proofs can be written in one of two ways: two columns, or a paragraph. A paragraph proof is only a two-column proof written in sentences. However, since it is easier to leave steps out when writing a paragraph proof, we'll learn the two-column method.

A two-column geometric proof consists of a list of statements, and the reasons that we know those statements are true. The statements are listed in a column on the left, and the reasons for which the statements can be made are listed in the right column. Every step of the proof (that is, every conclusion that is made) is a row in the two-column proof.

Writing a proof consists of a few different steps.

Draw the figure that illustrates what is to be proved. The figure may already be drawn for you, or you may have to draw it yourself.

List the given statements, and then list the conclusion to be proved. Now you have a beginning and an end to the proof.

Mark the figure according to what you can deduce about it from the information given. This is the step of the proof in which you actually find out how the proof is to be made, and whether or not you are able to prove what is asked. Congruent sides, angles, etc. should all be marked so that you can see for yourself what must be written in the proof to convince the reader that you are right in your conclusion.

Write the steps down carefully, without skipping even the simplest one. Some of the first steps are often the given statements (but not always), and the last step is the conclusion that you set out to prove. A sample proof looks like this:

Given:

Segment AD bisects segment BC.

Segment BC bisects segment AD.

Prove:

Triangles ABM and DCM are congruent.

Notice that when the SAS postulate was used, the numbers in parentheses correspond to the numbers of the statements in which each side and angle was shown to be congruent. Anytime it is helpful to refer to certain parts of a proof, you can include the numbers of the appropriate statements in parentheses after the reason.

3 0
3 years ago
Each of the following tables defines a relationship between an input x and an output y. Which of the relationship respresent fun
sveticcg [70]

Answer:

wat

Step-by-step explanation:

nose

6 0
3 years ago
Other questions:
  • I am confused on numbers 25 and 29, the instructions are at the top.
    13·1 answer
  • 1x + 1y= 2 , -3x + 4y= 15<br>using solving of elimination
    9·1 answer
  • Two groups of people entered a Drive-In restaurant. The first group ordered 6
    11·1 answer
  • PLEASE HELP ME FIND THE LENGTH
    8·2 answers
  • Simplify the expression 5/x^-2y^5
    11·1 answer
  • The fifth root of x to the fourth power times the fifth root of x to the fourth power
    7·2 answers
  • Which of these steps can be used to keep the equation balanced?
    14·1 answer
  • A)How much of the following shape is shaded?
    10·1 answer
  • FIND THE PRODUCTS OR QUOTIENYS IN THE EXPONENTIAL FORMS BY USING LAWS OF INDICES.​
    9·1 answer
  • What is the sum of 2 to the power of 3
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!