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NeTakaya
3 years ago
11

3(4x + 5) = 12 Which of the following correctly shows the beginning steps to solve this equation?

Mathematics
1 answer:
Contact [7]3 years ago
8 0
The correct answer is A.

Here is the completed problem:

3(4x+5)=12

Distributive property.

3(4x+5)=12

Simplify

12x+15=12

Subtract 15 from both sides.

12x+15−15=12−15

12x=−3

Divide both sides by 12.

12x/12 = −3/12

x= −1/4
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Topic Test
lara [203]

Answer:

B. P'(-1, 0) and Q'(1, 6)

Step-by-step explanation:

Please select the best answer from the choices provided

5 0
3 years ago
Read 2 more answers
(High points) please solve with explanation
AysviL [449]

Answer:

The area and the perimeter of the picture are:

  • <u>Area = 160 cm^2</u>
  • <u>Perimeter = 67.31 cm</u>

Step-by-step explanation:

To find the area of that figure, you can find the area how if it was a rectangle and next subtract the area of the triangle in the upper part. The area of a rectangle could be found by the next formula:

  • Area of a rectangle = base * height

As you can see in the picture, the base is 16 cm and the height is 12 cm, then we replace in the formula:

  • Area of a rectangle = 16 cm * 12 cm
  • Area of a rectangle = 192 cm^2

Now, we calculate the area of the triangle to subtract to the area we found and obtain the real area, the formula to obtain the area of a triangle is:

  • Area of a triangle = (base * height) / 2

The height of the triangle is 8 cm, and the base is 8 cm too, because you subtract to the base of the rectangle (16 cm) the measurements in the upper part (16 - 4 - 4 = 8), Now, we replace in the formula:

  • Area of a triangle = (8 cm * 8 cm) / 2
  • Area of a triangle = (64 cm^2) / 2
  • Area of a triangle = 32 cm^2

We subtract to the found area:

  • Area of the picture = 192 cm^2 - 32 cm^2
  • <u>Area of the picture = 160 cm^2</u>

To find the perimeter, you must add all the sides of the picture, but, as you can see, there is a side that doesn't have the measurent, this is the hypotenuse of the triangle used before, but how we know the other sides, we can use Pythagorean theorem:

  • a^{2}+b^{2}=c^{2}

Where:

  • a = Opposite leg (8 cm)
  • b = Adjacent leg (8 cm)

So, we replace in the theorem:

  • a^{2}+b^{2}=c^{2}
  • (8 cm)^{2}+(8cm)^{2}=c^{2} (and we clear c)
  • \sqrt{(8 cm)^{2}+(8cm)^{2}} =c
  • \sqrt{64 cm^{2}+64cm^{2}} =c
  • \sqrt{128cm^{2}} =c
  • c = 11.3137085 cm
  • c ≅ 11.31 cm

At last, we add all the sides of the picture begining by the base and going by the left side:

  • Perimeter of the picture = 16 cm + 12 cm + 4 cm + 11.31 cm + 8 cm + 4 cm + 12 cm
  • <u>Perimeter of the picture = 67.31 cm approximately</u>.
7 0
3 years ago
Type the correct answer in each box. Round the vector's magnitude to the nearest tenth.
Mama L [17]

Component form of u is (-18,13) and The magnitude of u is 22.2

<h3>What is a vector?</h3>

A vector is a two-dimensional entity with both magnitude and direction. A vector can be visualized geometrically as a directed line segment.

The component form of a vector is an ordered pair that describes the change in x and y values

This is mathematically expressed as (Δx, Δy) where Δx=x₂-x₁ and Δy=y₂-y₁

Given ;

Initial points of the vector as (14,-6)

The terminal point of the vector is (-4,7)

Here x₁=14,x₂=-4, y₁=-6,y₂=7

The component form of the vector u is (-4-14,7--6) =(-18,13)

Finding the Magnitude of the vector

u=√(x₂-x₁)²+(y₂-y₁)²

u=√-18²+13²

u=√324+169

u=√493

u=22.2

Therefore the Component form of u is (-18,13) and The magnitude of u is 22.2

To know more about vectors follow

brainly.com/question/25705666

#SPJ1

3 0
1 year ago
Two sides of a triangle measure 10 inches and 13 inches. The included angle between these sides is 55°. What is the approximate
PolarNik [594]
The answer should B 11.2 inches
4 0
3 years ago
Find the area of the rectangle with a length of (x^2-2) and a width of (2x^2-x+2)
Katyanochek1 [597]
Area=legnth times width

so multiply them together use distributive property
a(b+c)=ab+ac so
in this problem
(a+b)(c+d+e)=(a+b)(c)+(a+b)(d)+(a+b)(e)

so
x^2-2 times (2x^2-x+2)=(x^2)(2x^2-x+2)-(2)(2x^2-x+2)=(2x^4-x^3+2x^2)-(4x^2-2x+4)
add like terms
2x^4-x^3+(2x^2-4x^2)-2x+4
2x^4-x^3-2x^2-2x+4


5 0
3 years ago
Read 2 more answers
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