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3 years ago

23 + 4 + x = 25 -12 porfa ayuda :(

Mathematics

1 answer:

MAXImum [283]3 years ago

4 0

Answer:

x = - 14

Step-by-step explanation:

23 + 4 + x = 25 - 12

Combine like terms on both sides of the equation (by adding 23 and 4 on the left-hand side, and subtracting 12 from 25 on the right-hand side):

23 + 4 + x = 25 - 12

27 + x = 13

Next, subtract 27 from both sides to solve for x:

27 - 27 + x = 13 - 27

x = - 14

Please mark my answers as the Brainliest if you find this explanation helpful :)

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3. There were 567 whos in Whoville. Each who ate 814 bowls of

Rasek [7]

All together, there were 461,538 bowls of stew eaten all together.

6 0

3 years ago

4x – 4 = – 12 X = ? What is X ?

maxonik [38]

Answer:

X=-2

Step-by-step explanation:

4x-4=-12

Add 4 on both sides

+4 +4

4x=-8

Divide by 4 on both sides

÷4 ÷4

X=-2

8 0

3 years ago

Read 2 more answers

33, 25, 42, 25, 31, 37, 46,29,38 What is the first quartile of the data?

NISA [10]

answer:
Step by step solution:
1) put the numbers in order:
25,25,29,31,33,37,38,42,46
2) find the median:
25,25,29,31,33,37,38,42,46
33
3) find the median of the first quartile:
25,25,29,31
25 and 29
25+29= 54
54/2
27
Q1= 27

6 0

3 years ago

A can of tuna has a volume 18pi cm3 and a height of 2 cm. Find the area of the label that wraps around the entire can and does n

Sphinxa [80]

Answer:

The area of the label is $12\pi\ cm^{2}$

Step-by-step explanation:

step 1

Find the radius

we know that

The volume of the cylinder (can of tuna) is equal to

$V=\pi r^{2} h$

In this problem we have

$V=18\pi\ cm^{3}$

$h=2\ cm$

substitute and solve for r

$18\pi=\pi r^{2} (2)$

Simplify

$18=r^{2}(2)$

$r^{2}=9$

$r=3\ cm$

step 2

Find the lateral area of the can

The lateral area of the cylinder (can of tuna) is equal to

$LA=2\pi rh$

we have

$r=3\ cm$

$h=2\ cm$

$LA=2\pi (3)(2)=12\pi\ cm^{2}$

5 0

3 years ago

single die is rolled twice. Find the probability of rolling an oddodd number the first time and a number greater than 33 the sec

givi [52]

Answer:

$\frac{1}{4}$

Step-by-step explanation:

A single die is rolled twice, we have to find the probability of rolling an odd number in first throw and a number greater than 3 in the second throw.

a) Rolling an Odd number in first throw

A die has total 6 possible outcomes, out of which 3 are odd numbers i.e. 1,3 and 5

So, total number of possible outcomes = 6

Total Favorable outcomes (Odd numbers) = 3

Probability is defined as the ratio of favorable outcomes to total number of outcomes. So,

The probability of rolling an odd number would be = $\frac{3}{6} = \frac{1}{2}$

b) Rolling a number greater than 3 in second throw

Here again total possible outcomes = 6

Favorable outcomes (Numbers greater than 3 are 4, 5 and 6) = 3

So,

The probability of rolling a number greater than 3 = $\frac{3}{6} = \frac{1}{2}$

These two events(rolls) are independent of each other, so the overall probability of both events occurring would be the product of individual probabilities.

So,

Probability of rolling an odd number the first time and a number greater than 3 the second time = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

6 0

3 years ago