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2 years ago

Describe and correct the error in rewriting the equation

Mathematics

1 answer:

Sloan [31]2 years ago

5 0

2x - y = 5 <em>subtract 2x from both sides</em>

2x - 2x - y = 5 - 2x

- y = 5 - 2x

The sign "-" (minus) before y was lost.

You might be interested in

8 1/4 minus 5/6 equals what??

kondaur [170]

Ok to figure it out you first find the common factor, which is 12
so the fractions become 8 3/12 and 10/12 then you subtract those 2,
8 3/12-10/12 is 7 5/12.
Hope That Helps!!!

6 0

3 years ago

A rectangle has a length of 4x plus 5 and a width of 8x minus 4. Which expression shows the perimeter of that rectangle? A: 4x +

mezya [45]

Answer: The answer is (D) $24x+2.$

Step-by-step explanation: Given that there is a rectangle, where length of the rectangle is given by

$\ell=4x+5,$

and width of the rectangle is given by

$b=8x-4.$

We are to find the perimeter of the rectangle. To do so, we need to add two times length with two times breadth of the rectangle.

Therefore, perimeter is given by

$p=2\ell+2b=2\times(4x+5)+2\times(8x-1)=8x+10+16x-8=24x+2.$

Thus, the correct option is (D).

6 0

3 years ago

What is the length of the second base of a trapezoid if the length of one base is 24 and the length of the mid segment is 19

rodikova [14]

The sum of the two bases is twice the length of the mid segment
24+x=19*2
x=14

6 0

2 years ago

the head of a nail is circular , as shown the head of this nail has a diameter of 6 millimeters what measurment is closet to the

Nikitich [7]

Answer:

Area of nail's head = 28.26 millimeter² (Approx.)

Step-by-step explanation:

Given:

Head of nail is circular

Diameter of nail's head = 6 millimeter

Find:

Area of nail's head

Computation:

Radius of nail's head = Diameter / 2

Radius of nail's head = 6 / 2

Radius of nail's head = 3 millimeter

Area of circle = πr²

Area of nail's head = πr²

Area of nail's head = (22/7)(3)²

Area of nail's head = (22/7)(9)

Area of nail's head = (3.14)(9)

Area of nail's head = 28.26 millimeter² (Approx.)

5 0

2 years ago

A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set

lara [203]

A circle with center $(3,0)$ and radius $5$ has equation

$(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x = 16$

If we substitute $y=2x+k$ in this equation, we have

$x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0$

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

$\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0$

The expression simplifies to

$-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89$

The solutions to the associated equation are

$k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}$

So, the parabola is negative between the two solutions:

$-6-5\sqrt{5}$

8 0

3 years ago