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3 years ago

Solve the equation 3/4x-5=x

Mathematics

1 answer:

pav-90 [236]3 years ago

7 0

Answer:

x = -20

Step-by-step explanation:

3/4x-5=x

Subtract 3/4x from each side

3/4x - 3/4x -5=x-3/4x

3/4x - 3/4x -5=4/4x-3/4x

-5 = 1/4x

Multiply each side by 4

4 * -5 = 1/4x * 4

-20 =x

You might be interested in

A 2x2 square is centered on the origin. It is dilated by a factor of 3. What are the coordinated of the vertices of the square?

Alexxx [7]

<u>Answer</u>:

The vertices are:

A' = (-3, -3)

B' = (3, -3)

C' = (3, 3)

D' = (-3, 3)

The ratio of area of larger square to smaller square is 9:1

<u>Step-by-step explanation:</u>

Given:

A 2 x 2 square is centered at the origin.

So, the center of the square is (0, 0)

Since it is 2 x 2 square, the side of the square is 2 units.

So, the vertices of the 2 x 2 square are A (-1, -1), B(1, -1), C(1. 1), D(-1, 1)

The above square is dilated by a factor of 3.

Let's name the dilated square A'B'C'D'

To find the coordinates of the vertices of dilated square, we need to multiply each vertices of ABCD by 3.

A(-1, -1) = 3(-1, -1) = A'(-3, -3)

B(1, -1) = 3(1, -1) = B'(3, -3)

C(1, 1) = 3(1, 1) = C'(3, 3)

D(-1, 1) = 3(-1, 1) = D'(-3, 3)

To find the area of the small square

the side of the small square is 2 units

so the are of the small square is $2^2$ = 4 square units

To find the area of the larger square

lets find the side AB of the square using distance formula

=> $\sqrt{(x_2 -x_1)^2 +(y_2-y_1)^2}$

=> $\sqrt{(3 - (-3))^2 +(-3 - (-3))^2}$

=> $\sqrt{(3 +3)^2 +(-3 +3)^2}$

=> $\sqrt{(6)^2 +(0)^2}$

=> $\sqrt{36}$

=>6

AB =6 units

In a square all the sides will be equal

Now the area of the larger square will be

$6^2$

36 square units

The ratio of larger square to smaller square is

=>36 : 4

=>9 : 1

5 0

3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.