All categories

3 years ago

An equation that equals 19

Mathematics

2 answers:

olga nikolaevna [1]3 years ago

8 0

5 + 14 :) yyayayayayaya

taurus [48]3 years ago

8 0

Answer:

There are many different types of equations we can use to allow x to equal 19! Below are some examples :) See if you can come up with some on your own based off of these examples:

5x = 95.....................................Divide both sides b 5

x = 19.......................................There's one for you :)

1 + x - 5 = 15............................Combine like terms

x - 4 = 15..................................Add 4 to both sides

x = 19

You might be interested in

I attached an image. someone help me asap. i'm not the smartest person so i cont understand it :(

vampirchik [111]

Answer:

for the second column it is 12.5

for the third column it is 20%

for the fourth column it is 120%

for the fifth column it is 33%

for the sixth column it is 11.35

for the seventh column it is 900

for the 8th column it is 60%

for the final column it's 39

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Leto [7]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Using f (x) = 4x + 3 and g(x) = x - 2, f(g(5))=

il63 [147K]

Answer:

I think It is 7

Step-by-step explanation:

5 0

3 years ago

Kamden is a mind-reading alien. He had each friend draw and look at a card, then he guessed whether the card had a star on it.

klasskru [66]

Answer:

Step-by-step explanation:

corvet corvet

6 0

3 years ago

Newton's law of cooling is:

Mnenie [13.5K]

Answer:

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

$\frac{du}{dt}= -k (u-T)$

We can reorder the expression like this:

$\frac{du}{u-T} = -k dt$

We can use the substitution $w = u-T$ and $dw =du$ so then we have:

$\frac{dw}{w} =-k dt$

IF we integrate both sides we got:

$ln |w| = -kt +C$

If we apply exponential in both sides we got:

$w = e^{-kt} *e^c$

And if we replace w = u-T we got:

$u(t)= T + C_1 e^{-kt}$

We can also express the solution in the following terms:

$u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}$

For this case we know that $k =-0.15 hr$ since w ehave a cooloing, $T_{i}= 70 F, T_{amb}=11F$ , we have this model:

$u(t) = (70-11) e^{-0.15t} +11$

And if we want that the temperature would be 32F we can solve for t like this:

$32 = 59 e^{-0.15 t} +11$

$21=59 e^{-0.15 t}$

$\frac{21}{59} = e^{-0.15 t}$

If we apply natural logs on both sides we got:

$ln (\frac{21}{59}) =-0.15 t$

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

So it would takes approximately 6.9 hours to reach 32 F.

7 0

3 years ago