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otez555 [7]
2 years ago
14

Determine if the following expressions are equivalent. If so, by what property are they equivalent.

Mathematics
1 answer:
jeka942 years ago
5 0

Answer:

C. Yes, by distributive property

Step-by-step explanation:

Distributive property states that 3 can be distributed to all of the terms inside of the parentheses, namely x and -2. 3x + 3(-2) = 3x-6

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Does anyone understand this
melomori [17]
I hope this helps you

7 0
3 years ago
Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

5 0
3 years ago
A fruit vendor sells 85 pieces of fruit that are either apples or oranges. The ratio of apples to oranges is 3:2. How many apple
crimeas [40]
53. The ratio of apple over orange = 3:2 The total number of fruits are 60 pieces. Now, find the total number of apples and total number of oranges. => first, let’s add 2 + 3 = 5 Thus, we have 5 division for 60 pieces fruits, => 60 / 5 => 12 let’s solve for the apple => 12 x 3 = 36 Now. The orange => 12 * 2 = 24 => 36 + 24 = 60 Thus the ratio is 36 : 24
5 0
3 years ago
How many solutions does the equation −5a + 5a + 9 = 8 have?
faust18 [17]

Answer: the answer will be none

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513
dimaraw [331]

Answer:

Part 4) sin(\theta)=\frac{12}{13}

Part 10) The angle of elevation is 40.36\°

Part 11) The angle of depression is 78.61\°

Part 12) arcsin(0.5)=30\°  or arcsin(0.5)=150\°

Part 13) -45\°  or 225\°

Step-by-step explanation:

Part 4) we have that

cos(\theta)=-\frac{5}{13}

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

sin^{2}(\theta)+ cos^{2}(\theta)=1

substitute the given value

sin^{2}(\theta)+(-\frac{5}{13})^{2}=1

sin^{2}(\theta)+(\frac{25}{169})=1

sin^{2}(\theta)=1-(\frac{25}{169})  

sin^{2}(\theta)=(\frac{144}{169})

sin(\theta)=\frac{12}{13}

Part 10)

Let

\theta ----> angle of elevation

we know that

tan(\theta)=\frac{85}{100} ----> opposite side angle theta divided by adjacent side angle theta

\theta=arctan(\frac{85}{100})=40.36\°

Part 11)

Let

\theta ----> angle of depression

we know that

sin(\theta)=\frac{5,389-2,405}{3,044} ----> opposite side angle theta divided by hypotenuse

sin(\theta)=\frac{2,984}{3,044}

\theta=arcsin(\frac{2,984}{3,044})=78.61\°

Part 12) What is the exact value of arcsin(0.5)?

Remember that

sin(30\°)=0.5

therefore

arcsin(0.5) -----> has two solutions

arcsin(0.5)=30\° ----> I Quadrant

or

arcsin(0.5)=180\°-30\°=150\° ----> II Quadrant

Part 13) What is the exact value of arcsin(-\frac{\sqrt{2}}{2})

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

sin(45\°)=\frac{\sqrt{2}}{2}

therefore

arcsin(-\frac{\sqrt{2}}{2}) ----> has two solutions

arcsin(-\frac{\sqrt{2}}{2})=-45\° ----> IV Quadrant

or

arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\° ----> III Quadrant

5 0
3 years ago
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