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artcher [175]
3 years ago
14

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

Mathematics
2 answers:
harina [27]3 years ago
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


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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
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