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4 years ago

Ayyyyyyyyyyyyyyyy i need some help lease

Mathematics

1 answer:

ki77a [65]4 years ago

6 0

Answer:

.95

Step-by-step explanation:

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A jug has a maximum capacity of 14 liters. is filled to 60% of its capacity, and then poured out at a rate of 60 mL/s. Which of

ZanzabumX [31]

Answer:

W(x) = 3.4 L - (25/1000)(L/s)*x

Step-by-step explanation:

it a tip

W(x) = 3.4 L - (25/1000)(L/s)*x

The maximum of the jug is 4 L.

now, the jug is 85% filled, then the amount of water in the jug is:

(85%/100%)*4L = 0.85*4L = 3.4L

Now, we pour 25 mL each second, so we could write a linear equation as:

W(x) = 3.4L - (25ml/s)*x

where x is the number of seconds that passed since the beginning,

But we want to write this in Liters, we have that:

1L = 1000mL.

Then 1mL = (1/1000) L

then we can write 25 mL as:

25m L = 25*(1/1000) L = (25/1000) L

Then we can write the equation as

W(x) = 3.4 L - (25/1000)(L/s)*x

which is the amount of water remaining in the jug after x seconds.

7 0

3 years ago

Evalulate 3*5-2*(3+2)

likoan [24]

Answer:

Step-by-step explanation:

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

Following the order, first is parenthesis:

3 · 5 - 2(5)

Multiplication:

15 - 10

8 0

3 years ago

Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx

Rudiy27

If we use the substitution $u = \tan x$ , then $du = \sec^2 {x}\ dx$ . If you try substituting just $u$ and $du$ into the integrand, though, you'll notice that there's a $\sec^2x$ left over that we have to deal with.

To get rid of this problem, use the identity $\tan^2 x + 1 = \sec^2 x$ and substitute in the left side of the identity for the extra $\sec^2x$ , as shown:

$\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx$
$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$

From there, we can substitute in $u$ and $du$ , and then evaluate:

$\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du$
$\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du$
$= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}$
$= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})$
$= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}$

8 0

3 years ago

The coldest place in the Milky Way galaxy is found in the Boomerang Nebula. The temperature is -457.6°F. Convert this to a Celsi

valentina_108 [34]

-272 celsius is the answer

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3 years ago

Read 2 more answers

Sandra baked 36 cookies yesterday morning and 24 cookies last night. Tomorrow, she plans on baking 48 more. How many cookies did

pshichka [43]

Answer B) 60 because 34 + 24 is 60

3 0

3 years ago