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3 years ago

The temperature at midnight is -5°C and is expected to drop 12°

Mathematics

1 answer:

victus00 [196]3 years ago

8 0

Answer:

-17 degrees Celsius.

Step-by-step explanation:

If the temperature at midnight is -5, and it is supposed to drop 12 more by sunrise, the 12 degrees are subtracted from the initial temperature.

-5 - 12 = -17 degrees by sunrise.

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Which choices are equivalent to the expression below? Check all that apply.

Gre4nikov [31]

Answer:

A and E

Step-by-step explanation:

Sqrt 3 times sqrt 3 is sqrt 9, which simplified is 3

3 0

2 years ago

To evaluate an expression with variables: 1. Substitute the values for the variables. 2. Identify grouping symbols, beginning wi

Eduardwww [97]

Evaluate the expression using x = –1 and y=3.

$(6x^4y^3)$

1. Substitute the values for the variables.

we plug in -1 for x and 3 for y

$(6x^4y^3)$

$(6(-1)^4(3)^3)$

2. Identify grouping symbols, beginning with the innermost symbols.

$(6*(-1)^4*(3)^3)$

We take exponent and then we multiply

3. Determine and apply the rules of exponents to simplify

$(6*(-1)^4*(3)^3)$

(-1)^4 = -1 * -1 * -1 * -1 = 1

(3)^3 = 3*3*3= 27

$(6*(1)*(27))$

4. Evaluate.

$(6*(1)*(27))$

162

The value of the expression is 162

6 0

3 years ago

Read 2 more answers

Please help with this calculus problem!

ELEN [110]

$\dfrac{\mathrm dQ}{\mathrm dt}=\dfrac{a(1-5bt^2)}{(1+bt^2)^4}$
$Q(t)=\displaystyle\int\frac{a(1-5bt^2)}{(1+bt^2)^4}\,\mathrm dt$

Let $t=\dfrac1{\sqrt b}\tan u$ , so that $\mathrm dt=\dfrac1{\sqrt b}\sec^2u\,\mathrm du$ . Then the integral becomes

$\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{(1+\tan^2u)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{\sec^6u}\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du$

One way to proceed from here is to use the power reduction formula for cosine:

$\cos^{2n}x=\left(\dfrac{1+\cos2x}2\right)^n$

You'll end up with

$=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du$
$=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C$
$=\dfrac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C$
$Q(t)=\dfrac{at}{(1+bt^2)^3}+C$

6 0

3 years ago

Ann is growing two different plants for a science project. Plant A grew 4/12 inch the first week and 2/12 inch the second week.

Galina-37 [17]

Start the number line from 0 and put divisions at 1/12. Then mark at 4/12 for plant A and indicate this was its growth in the first week. Then, make another mark up to 6/12 and indicate this was the plant's growth in the second week.
For plant B, make one mark at 7/12 and indicate that this is plant B's growth in the first week.

After two weeks, plant A has a height of 6/12 inches and plant B has a height of 7/12 inches, so B is taller.

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3 years ago

What is 3 divided by 31

Basile [38]

0.09677419354 = 3 divided by 31

8 0

3 years ago

Read 2 more answers