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3 years ago

Slove:33/29=46/x. Try to find the answer.

Mathematics

1 answer:

Tju [1.3M]3 years ago

5 0

Answer:

$x=40\frac{14}{33}$

Step-by-step explanation:

we have

$\frac{33}{29}=\frac{46}{x}$

Solve for x

Multiply in cross

$(33)(x)=(46)(29)$

Divide by 33 both sides (Division Property of Equality)

$x=46(29)/33$

$x=\frac{1,334}{33}$

Convert to mixed number

$x=\frac{1,334}{33}=\frac{1,320}{33}+\frac{14}{33}=40\frac{14}{33}$

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S = ut+ 16t²<br>Find the value of s when u = 2 and t = 3.

Fed [463]

Answer:

Step-by-step explanation:

Given,

u = 2

t = 3

Now, Finding the value of s

$s = ut \: + 16 {t}^{2}$

Plugging the values of u and t

$s = 2 \times 3 + 16 \times {( 3)}^{2}$

Calculate

$s = 2 \times 3 + 16 \times 9$

$s = 6 + 144$

Calculate the sum

$s = 150$

Hope this helps ..

Good luck on your assignment..

4 0

3 years ago

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Evaluate k(x) = -k - 7 when k(-4)

FinnZ [79.3K]

Answer:

I'm not sure but i think that it's

x= -7

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8 0

3 years ago

BRAINLIESTTTT ASAP! PLEASE ANSWER

iren [92.7K]

H(t) = −16t^2 + 75t + 25
g(t) = 5 + 5.2t

A)
At 2, h(t) = 111, g(t) = 15.4
At 3, h(t) = 106, g(t) = 20.6
At 4, h(t) = 69, g(t) = 25.8
At 5, h(t) = 0, g(t) = 31
The heights of both functions would have been the closest value to each other after 4 seconds, but before 5 seconds. This is when g(x) is near 30 (26-31), and the only interval that h(t) could be near 30 is between 4 and 5 seconds (as it is decreasing from 69-0).

B) The solution to the two functions is between 4 and 5 seconds, as that is when their height is the same for both g(t) and h(t). Actually the height is at 4.63 seconds, their heights are both
What this actually means is that this time and height is when the balls could collide; or they would have hit each other, given the same 3-dimensional (z-axis) coordinate in reality.

6 0

3 years ago

Read 2 more answers

I will mark as the brainliest answer <br><br> Please show your work

swat32

3 and 5 are the answer

4 0

3 years ago

Suppose that poaching reduces the population of an endangered animal by 6% per year. Further suppose that when the population o

artcher [175]

If poaching reduces the population of an endangered animal by 6% and the criteria of population extinction is 20 with present population 1500 then it will take 3.62 years to reach to the mark of extinction.

Given Poaching reduces the population of endangered animals by 6% per year. The criteria of population extinction is 20. Present population being 1500.

Number of years taken by the population of endangered animals to reach to 20 mark can be calculated as under:

20=1500* $(1-0.06)^{n}$

20=1500* $0.94^{n}$

20/1500= $0.94^{n}$

0.133= $0.94^{n}$

take log both sides

log(0.133)=log $0.94^{n}$ -------1

log(0.133)=nlog (0.94)

put the values log values:

log(0.133)=-0.8761

log(0.94)=-0.0268

Taking 1

-0.8761=n*(-0.0268)

n=0.8761/0.0268

n=3.269

Hence to reach level of 20 the population takes 3.2 years.

Learn more about logarithm at brainly.com/question/25710806

#SPJ10

8 0

2 years ago