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4 years ago

13

If the overnight temperature at the Arctic Circle was -14 degrees, but the temperature rose to 9 degrees during the day, what wa

s the difference between the high and low temperatures?

1 answer:

Tpy6a [65]4 years ago

7 0

-5 is the answer. -14 + 9= -5

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How do you write 21% as a decimal

Goshia [24]

Answer:

0.21

Step-by-step explanation:

if you write it as a fraction it would be 21/100

and if that is made to a decimal it would be 0.21 .

8 0

3 years ago

Read 2 more answers

(PLEASE ANSWER ASAP) Which number is between 1/6 and 0.25?

serg [7]

If you change both to a fraction you would see that you have 1/6 and 1/4 (.25 as a fraction). So 1/5 would be in between them.

You could also change both as a decimal. 1/6 =.166666 and .25 so any number in between them would be correct such as .2 which is the same as 1/5.

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4 years ago

Help me quickly! NO LINKS!!!!

AURORKA [14]

Answer:

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3 years ago

Read 2 more answers

taurus [48]

Answer:

$x^2 - 2x + 10$ , option B

Step-by-step explanation:

Complex numbers:

The most important relation that involves complex numbers is given by:

$i^2 = -1$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

The solutions are:

$x_1 = 1 - 3i, x_2 = 1 + 3i$

We have to find the polynomial. All option have $a = 1$ . So

$(x - (1 - 3i))(x - (1 + 3i)) = x^2 - x(1 + 3i) - x(1 - 3i) + (1 - 3i)(1 + 3i) = x^2 - x -3ix - x + 3ix + 1^2 - (3i^2) = x^2 - 2x + 1 - 9i^2 = x^2 - 2x + 1 - 9(-1) = x^2 - 2x + 10$

The correct answer is given by option b.

6 0

3 years ago

At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while

aliya0001 [1]

Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is $\theta$

Thus, by the question,

$tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}$

$\implies tan\theta = \frac{x}{x}$ ( at 3pm the height of shadow = height of the pole)

$\implies tan\theta = 1$

$\implies \theta = 45^{\circ}$

Again, according to the question,

When the angle of elevation is $(\theta - 12^{\circ})$ ,

The height of shadow = x + 95,

$\implies tan(45-12)^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x$

$\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}$

$\implies -0.3505924068 x = - 61.6937213538$

$x=175.969930202\approx 175.97\text{ cm}$

4 0

3 years ago