All categories

3 years ago

If 9 is subtracted from two thirds of a number , the result is 7. find the number.

2 answers:

7 0

So what you do is (2/3)x-9=7. Add 9 to both sides (2/3)x=16. Multiply by reciprocal on both sides 16 x (3/2). 16 x 3 = 48. 48/2=24

Mrac [35]3 years ago

5 0

24 it's 24, Yes 24 is big brain answer :)

You might be interested in

Find the surface area of a cylinder with a height of 7 m and a base radius of 6 m.

weqwewe [10]

Answer:

489.84 m²

Step-by-step explanation:

Area of one 2d circle: πr² ⇒ π6² ⇒ 36π ≈ 113.04 (using 3.14 for pi)

Area of both 2d circles: 113.04 + 113.04 =226.08 m²

Now we have to find the width of the rectangle, which is equal to the circumference of either circle:

Width of rectangle: 2πr ⇒ 2π6 = 12π ≈ 37.68 (using 3.14 for pi)

We can find the area of the rectangle now, since the length was given

Area of rectangle: 37.68· 7= 263.76 m²

Surface Area: 263.76+226.08= 489.84m²

Hopefully this helps!

6 0

3 years ago

I need help I don't understand it.

BartSMP [9]

These angles are corresponding angles, and we know that corresponding angles are equal when a transversal intersects two parallel lines.

So,

8x - 12 = 6x + 8

=> 8x - 6x = 8 + 12

=> 2x = 20

=> x = 20/2

=> x = 10

So, the value of x is 10.

3 0

2 years ago

Read 2 more answers

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

3 years ago

Is the data set 3,3,3,4,4,5,25 skewed right skewed left or normally distributed?

OverLord2011 [107]

It is skewed to the right because 25 is the outlier.

3 0

3 years ago

Gggrrgggjgfgg fr kk had

notka56 [123]

Answer:

B) No

Step-by-step explanation:

the expressions are not equivalent

6 0

2 years ago

Read 2 more answers