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3 years ago

Hat term can you add to StartFraction 5 over 6 EndFraction x minus 4 to make it equivalent to One-half x minus 4?

Mathematics

1 answer:

kolezko [41]3 years ago

8 0

Answer:

$\dfrac{-x}{3}$ should be added to the given equation.

Step-by-step explanation:

Let a is added to the given equation.

According to the given condition,

$\dfrac{5x}{6}-4+a=\dfrac{x}{2}-4\\\\a=\dfrac{x}{2}-4+4-\dfrac{5x}{6}\\\\a=\dfrac{x}{2}-\dfrac{5x}{6}\\\\a=\dfrac{3x-5x}{6}\\\\a=\dfrac{-2x}{6}\\\\a=\dfrac{-x}{3}$

So, $\dfrac{-x}{3}$ should be added to the given equation.

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3 years ago

A square park measures 170 feet along each side. Two paved paths run from each corner to the opposite corner and extend 3 feet i

Cerrena [4.2K]

Answer:

The total area, in square feet, taken by the paths is 2,004

Step-by-step explanation:

see the attached figure with lines to better understand the problem

I can divide the figure into four right triangles, one small square and four rectangles

step 1

Find the area of the right triangle of each corner of the path

The area of the triangle is

$A=(1/2)(b)(h)$

substitute the given values

$A=(1/2)(3)(3)=4.5\ ft^2$

step 2

Find the hypotenuse of the right triangle

Applying Pythagoras Theorem

Let

d -----> hypotenuse of the right triangle

$d^{2}=3^{2}+3^{2}$

$d^{2}=18$

$d=\sqrt{18}\ ft$

simplify

$d=3\sqrt{2}\ ft$

The hypotenuse of the right triangle is equal to the width of the path

step 2

Find the area of the small square of the path

The area is

$A=b^{2}$

we have

$b=3\sqrt{2}\ ft$ ----> the width of the path

substitute

$A=(3\sqrt{2})^{2}$

$A=18\ ft^2$

step 3

Find the length of the diagonal of the square park

Applying Pythagoras Theorem

Let

D -----> diagonal of the square park

$D^{2}=170^{2}+170^{2}$

$D^{2}=57,800$

$D=\sqrt{57,800}\ ft$

simplify

$D=170\sqrt{2}\ ft$

step 4

Find the height of each right triangle on each corner

The height will be equal to the width of the path divided by two, because is a 45-90-45 right triangle

$h=1.5\sqrt{2}\ ft$

step 5

Find the area of each rectangle of the path

The area of rectangle is $A=LW$

we have

$W=3\sqrt{2}\ ft$ ----> width of the path

Find the length of each rectangle of the path

$L=(D-2h-d)/2$

where

D is the diagonal of the park

h is the height of the right triangle in the corner

d is the width of the path (length side of the small square of the path)

substitute the values

$L=(170\sqrt{2}-2(1.5\sqrt{2})-3\sqrt{2})/2$

$L=(170\sqrt{2}-3\sqrt{2}-3\sqrt{2})/2$

$L=(164\sqrt{2})/2$

$L=82\sqrt{2}\ ft$

Find the area of each rectangle of the path

$A=LW$

we have

$W=3\sqrt{2}\ ft$

$L=82\sqrt{2}\ ft$

substitute

$A=(82\sqrt{2})(3\sqrt{2})$

$A=492\ ft^2$

step 6

Find the area of the paths

Remember

The total area of the paths is equal to the area of four right triangles, one small square and four rectangles

substitute

$A=4(4.5)+18+4(492)=2,004\ ft^2$

therefore

The total area, in square feet, taken by the paths is 2,004

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3 years ago