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2 years ago

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The temperature at a mountain base camp was −3 degrees Celsius on Monday. Tuesday morning, the temperature was 2 degrees Celsius

lower than it was on Monday. By Tuesday evening, the temperature was 4 degrees Celsius lower than it was that morning. What was the temperature at the base camp Tuesday evening?

2 answers:

kotegsom [21]2 years ago

3 0

It is -3 to start off with, on Monday. On Tuesday morning it would be -3 - 2 degrees, = -5 degrees. By Tuesday evening it was 4 degrees lower than in the morning, so -5 - 4 degrees = -9 degrees on Tuesday evening.

Hope this helps xox :)

Arlecino [84]2 years ago

3 0

Answer:

The temperature was -9°C on Tuesday evening.

Step-by-step explanation:

Given,

The temperature on monday = -3°C,

Also, on tuesday morning, the temperature was 2 degrees Celsius lower than it was on Monday,

So, the temperature on tuesday morning = -3°C - 2°C = -5°C,

And, on tuesday evening, the temperature was 4 degrees Celsius lower than it was that morning,

Hence, the temperature on tuesday evening = -5°C - 4°C = -9°C

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The width of a rectangle is 6 kilometers less than twice its length. if its area is 108 square kilometers, find the dimensions

Art [367]

Hi there!

Answer:
length = 9 kilometres
Width = 12 kilometres

Let's solve this problem step by step!
To find our answer we need to set up and solve an equation.

Let the length of the rectangle be represented by x.
The width of the rectangle can therefore be expressed by 2x - 6.

The area of a rectangle can be found by using the formula:
A = width × length

Plug in the data from the formula
A = x (2x - 6).

Simplify using rainbow technique.
$x(2x - 6) = 2 {x}^{2} - 6x$

Now we've found the simplified expression that expresses the area of the rectangle. Therefore, we can now set up and start solve the equation.

$2 {x}^{2} - 6x = 108$
Subtract 108

$2 {x}^{2} - 6x - 108 = 0$
Divide by 2.

${x}^{2} - 3x - 54$
$(x - 9)(x + 6) = 0$
Rule AB = 0, gives A is 0 or B is 0.

$x - 9 = 0 \\ x = 9 \\ \\ x + 6 = 0 \\ x = - 6$

The length of the rectangle, which was represented by x, must be 9 (since it cannot be a negative number).

Length
$x = 9$
Width
$2x - 6 = 2 \times 9- 6 = 18 - 6 = 12$

Answer:
length = 9 kilometres
Width = 12 kilometres

~ Hope this helps you!

6 0

3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

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Answer:

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