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LuckyWell [14K]

2 years ago

7

The low temperature for 5 days was -5 degrees, -7 degrees, -2 degrees, 2 degrees, and - 3 degrees. What was the average temperat

ure for those days

1 answer:

weqwewe [10]2 years ago

5 0

The average low is -15 is that what you meant?

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3x(x-2)-x^2+4=0<br> Help

antoniya [11.8K]

Answer:

x=2,1 solving with quadratic formula

Step-by-step explanation:

3 0

2 years ago

What is equivalent to 6 + 3 x 9 - 4

STALIN [3.7K]

Answer:

29

Step-by-step explanation:

6 + 3 x 9 - 4

6 + 27 - 4

33 - 4

=29

3 0

2 years ago

647.6 ÷ what = 0.6476

pshichka [43]

Answer:

.001

Step-by-step explanation:

You see it is the same numbers just it was moved 3 places to the left.

So the you would multiply it by .001 <------ there are 3 decimal places.

When you multiply them you get.

647.6*.001= .6476

3 0

2 years ago

Read 2 more answers

Each month, edward budgets $1860 for fixed expenses, $860 for living expenses and $250 for annual expenses. His annual net incom

Vlad [161]

Edward's monthly expenses total 2970. Edward's monthly income amounts to 2709, an amount that is 261 less than his expenses.

The best description of Edward's monthly budget is that ...
It shows a deficit of $261.

3 0

2 years ago

Make 'h' as the subject. Remember to rationalize the denominator while solving.

Sliva [168]

Hello there BeGuiLE !

To solve your question, we must first bring 'h' to one side of the equation. Let's bring all the terms with the variable 'h' to the left side of the equation. So,
$\large{\sf\sqrt{ 3 } h = 1.6+h}$
→ $\large{\sf\sqrt{3}h-h=1.6 }$

Now, let's combine the 2 terms containing the variable 'h'.
$\large{\sf\sqrt{3}h-h=1.6 }$
→ $\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$

Now, we need to leave the variable 'h' alone in the left side of the equation & bring (√3 + 1) to the other side of the equation. This is done in order to make 'h' as the subject of the equation. So, we'll get it as,
$\large{\sf\left(\sqrt{3}-1\right)h=1.6 }$
→ $\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$

Now, let's rationalize it. Rationalizing is a process where we move the root from the denominator of a fraction to the numerator for easier calculation. So,
$\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1 )(\sqrt{3}+1}) }$
Now, use the algebraic identity, (a + b)(a - b) = a² - b²
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{3-1} }$
→ $\large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{2} }$
→ $\boxed{\large{\sf\:h=0.8(\sqrt{3}+1)}}$

Now, we've made 'h' as the subject of the equation. If you want to solve it more, then,
$\large{\sf\:h=0.8(\sqrt{3}+1)}$
Take √3 as 1.73 (approx. value up to 2 decimal places)
→ $\large{\sf\:h=0.8(1.73+1)}$
→ $\large{\sf\:h=0.8(2.73)}$
→ $\boxed{\boxed{\huge{\bf{\:h=2.184 \: (approx.)}}}}$

I hope this will help you.

Please refer to the attached image if the explanation shows some error.

_______

Check out more links which will help you understand the topic better :

■ brainly.com/question/21406377

■ brainly.com/question/696184

_______
$\mathfrak{Lucazz}$

4 0

2 years ago

Read 2 more answers