All categories

2 years ago

Solve for x. 8x^2−3=2

Mathematics

1 answer:

sergeinik [125]2 years ago

5 0

8x^2=2+3

8x^2=5

x^2=5/8

x= sqrt(5/8)

You might be interested in

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

2 years ago

Find the slope: a line passes through P (3,-7) and Q (-5,5)

nalin [4]

76 mutiply by each other

7 0

2 years ago

How do I solve inequalities with fractions on both sides?

poizon [28]

In this situation you would add 9 to the other side of the inequality 2/5x < 9/10+9
2/5x< 9.9
:)

4 0

2 years ago

The letters of the word S U M M E R were placed in a bag. What is the probability of choosing an R?

lawyer [7]

｡☆✼★ ━━━━━━━━━━━━━━ ☾

There is only one R in the bag

So the probability would be 1/6

1/6 as a decimal is 0.16

Thus, your answer would be option D

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━ ☾

3 0

2 years ago

Find the base circumference of a cone with height 10 cm and volume 90π cm3 .

Leni [432]

Answer:

Answer is circumference =32.64

Step-by-step explanation:

$given \\ volume \: of \: cone \: = 90\pi \: {cm}^{3} \\ and \: height = 10 \: cm \: \\ to \: find: circumference \\ by \: the \: given \: data \\ volume \: of \: cone = 90\pi \\ \frac{1}{3} \pi {r}^{2} h = 90\pi \\ cancelling \: \pi \: which \: is \: common \: \\ we \: get \\ \frac{1}{3 } {r}^{2} h = 90 \\ we \: know \: that \: h = 10 \\ on \: substituting \: we \: get \\ \frac{1}{3} {r}^{2} \times 10 = 90 \\ dividing \: by \: 10 \: on \: both \: sides \: we \: get \\ \frac{1}{3} {r}^{2} = 9 \\ {r }^{2} = 27 \\ r = 3 \sqrt{3} \\ we \: know \: that \: circumference \: = 2\pi \: r \\ = 2 \times \frac{22}{7} \times 3 \sqrt{3} \\ on \: solving \: the \: above \: mentioned \: we \: get \\ \\ \: = 32.64$

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

3 0

3 years ago

Read 2 more answers