All categories

3 years ago

Solve: negative 1 over 2 x + 2 = −x + 7 10 5 over 2 2 over 5 20

2 answers:

8 0

Rewrite the equation in mathematical form. That is,
- (1/2)x + 2 = -x + 7
Multiply both sides of the equation by 2 to get rid of the fraction,
- x + 4 = -2x + 14
Group all the terms with x and the constants on different sides of the equation,
-x + 2x = 14 - 4
Therefore, the value of x is 10.

yan [13]3 years ago

7 0

-1/2x + 2 = -x + 7
-1/2x + x = 7 - 2
1/2x = 5
x = 10

You might be interested in

From a circular cylinder of diameter 10 cm and height 12 cm are conical cavity of the same base radius and of the same height is

Nataliya [291]

<h3>Volume of the remaining solid = 628 cm^2</h3>

<h3>Whole surface area = 659.4 cm^2</h3>

Step-by-step explanation:

Now, Given that:-

Diameter (d) = 10 cm

So, Radius (r) = 10/2 = 5cm

Height of the cylinder = 12cm.

$volume \: of \: the \: cylinder \: = \pi {r}^{2} h$

$= > \pi \times {5}^{2} \times 12 {cm}^{3} = 300\pi {cm}^{3}$

Radius of the cone = 5 cm.

Height of the cone = 12 cm.

$slant \: height \: of \: the \: cone \: = \sqrt{ {h}^{2} + \: {r}^{2} }$

$= > \sqrt{ {5}^{2}+{12}^{2} } cm \: = 13cm$

Volume of the cone = 1/3 *πr^2h

$= > \frac{1}{3} \pi \times {5}^{2} \times 12 {cm}^{3} = 100\pi {cm}^{3}$

therefore, the volume of the remaining solid

$= 300\pi {cm}^{3} - 100\pi {cm}^{3} \\ = 200 \times 3.14 {cm}^{3} = 628 {cm}^{3}$

Curved surface of the cylinder =

$2\pi \: rh \: = 2\pi \times 5 \times 12 {cm}^{2} \\ = 120\pi {cm}^{2} .$

$curved \: surface \: of \: the \: cone \: = \pi \: rl \\ = \pi \times 5 \times 13 {cm}^{2} \\ = 65\pi {cm }^{2} \\ area \: of \: (upper)circular \: base \: \\ of \: cylinder \: = \\ = \pi \: {r}^{2} = \pi \times {5}^{2}$

therefore, The whole surface area of the remaining solid

= curved surface area of cylinder + curved surface area of cone + area of (upper) circular base of cylinder

$= 120\pi {cm}^{2} + 65\pi {cm }^{2} + 25 \pi {cm}^{2} \\ = 210 \times 3.14 {cm}^{2} = 659.4 {cm}^{2}$

<h3>Hope it helps you!!</h3>

6 0

2 years ago

Write the equation of each graph after the indicated transformations. The graph of y = √x is reflected in the x-axis, stretched

Zarrin [17]

Answer:

can you mark me of a brainliest please answers to you is C

7 0

3 years ago

Study the following data set.

Katen [24]

The interquartile range of the data set is 8

<h3>How to solve for the interquartile range</h3>

To do this we have to arrange in ascending order

= 8,9,9,9,10,11,13,15,17,18,22

Q2 = The median of the data is 11.

We have to find the median of the half before 11 and the median of the half after 11.

We would have

8,9,9,9,10

Q1 = 9

And 13,15,17,18,22

Q3 = 17

The interquartile range is

Q3 - Q1

= 17 - 9

= 8

The interquartile range of the data set is 8