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3 years ago

It was -5 celsius in Copenhagen and -12 celsius in Oslo. Which city was colder?

2 answers:

5 0

Since -12 is higher than -5, -5 would be a little bit more warmer than -12 and you want the coldest city so it would be Oslo

BARSIC [14]3 years ago

4 0

Answer:

Oslo

Step-by-step explanation: -12 is below -5. Therefore it is colder the further below on the temperature

You might be interested in

Wrich expreson is equivalent to 4(3-6).

Korvikt [17]

You haven’t provided any option, however I think I can still help.

You can solve 4(3-6). You first deal with he parenthesis, and get 4(-3). This means that your expression is equivalent to -12.

Hope this helps :)

8 0

3 years ago

The quotient of 49 and d increased by the product of k and f?

stich3 [128]

"d increased by the product of k and f" is d+kf

49/(d+kf)

The final answer is 49/(d+kf)~

6 0

3 years ago

Number line for 235+123

valentina_108 [34]

I think this is right
all you need to do is:

235-123= 112
then to make sure its correct you do:
123+112= 235

Hope this has helped you. :)

8 0

3 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

==========================================================

Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$

Answer is choice D

5 0

2 years ago

Read 2 more answers

Can someone answer this soon?!?

Sever21 [200]

First get all the terms on the same side:

$x^2 + 7x - 4 = 0$

Now solve using the quadratic formula which is:

$\frac{-b±\sqrt{b^2 - 4ac} }{2a}$

Plug in the values:

$\frac{-7± \sqrt{49-4(1)(-4)} }{2}$

So the answer is:

$\frac{-7 ± \sqrt{65} }{2}$

6 0

3 years ago