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2 years ago

In what temperature range can mercury be found as a liquid

1 answer:

4 0

Some metals melt at lower temps than others. Tin becomes liquid at 231.8˚C (447.8˚F); at the other end of the spectrum, tungsten has the highest melting point of any metal: 3,422˚C (6,192˚F). Somewhere in the middle is mercury, which stays in a liquid state until its temperature drops to -40.

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Transform the given quadratic function into vertex form f(x) = quadratic function into vertex form f(x) = quadratic function int

melisa1 [442]

Answer:

Vertex form: $f(x) =(x +\frac{7}{2})^2 -\frac{53}{4}$

The vertex is $(-\frac{7}{2},-\frac{53}{4})$

Step-by-step explanation:

For a general quadratic function the form is:

$ax ^ 2 + bx + c$

For the function

$f(x) = x ^ 2+ 7x -1$

The values of the coefficients for the function are the following: $a = 1$ , $b =7$ , $c = -1$

Take the value of b and divide it by 2. Then, the result obtained squares it.

$\frac{b}{2}= \frac{7}{2}$

$(\frac{b}{2})^2=(\frac{7}{2})^2=\frac{49}{4}$

Add and subtract $\frac{49}{4}$

$f(x) = (x ^ 2 +7x +\frac{49}{4}) -\frac{49}{4}- 1$

Write the expression of the form

$f(x) = (x+\frac{b}{2})^2 +k$

$f(x) =(x +\frac{7}{2})^2 -\frac{53}{4}$

The vertex is $(-\frac{7}{2},-\frac{53}{4})$

6 0

3 years ago

Pls help me I rlly don’t get it :(

zhenek [66]

Step-by-step explanation:

Use SOH-CAH-TOA.

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

Let's start with #12. The hypotenuse is 18. The side adjacent to ∠B is 6. Since we have the adjacent side and hypotenuse, we should use cosine.

cos B = 6/18

Solving for B:

B = cos⁻¹(6/18)

Using a calculator:

B ≈ 70.5°

Now let's do #14. The side adjacent to ∠B is 19, and the side opposite of ∠B is 22. Since we have the adjacent side and opposite side, we should use tangent.

tan B = 22/19

Solving for B:

B = tan⁻¹(22/19)

Using a calculator:

B ≈ 49.2°

8 0

2 years ago

I really need help with this, ill give brainly.

rusak2 [61]

Answer:

the rate of change is-3 2

3 0

2 years ago

What is the discontinuity and zero of the function f(x) = 3x^2 + x - 4 / x-1

7nadin3 [17]

Ans: Option (1) <span>Discontinuity at (1, 7), zero at ( negative four thirds , 0)
</span>
Explanation:
Given function:
$f(x) = \frac{3x^2 + x - 4}{x-1}$

Now if we plug in the x = 1, we would have discontinuity as function goes to infinity.

Now for f(1):
$f(x) = \frac{3x^2 + x - 4}{x-1} \\ f(x) = \frac{3x^2 + 4x - 3x - 4}{x-1} \\ f(x) = \frac{x(3x+4)-1(3x+4)}{x-1} \\ f(x) = \frac{(x-1)(3x+4)}{(x-1)} \\ f(x) = 3x+4 \\ now ~ insert ~ x=1: \\ f(1) = 3(1) + 4 = 7 \\ It~means~discontinuity~at~(1,7).\\ Now~let~us~find~zeros.\\ put~ f(x) = 3x+4 ~ equals~ to ~zero. \\ =\ \textgreater \ ~3x+4 = 0 \\ =\ \textgreater \ ~ x = \frac{-4}{3} \\ Hence~zeros=( \frac{-4}{3}, 0)$

5 0

3 years ago

The equation of line AB is y= 5x+1 . Write an equation of a line parallel to line AB in slope intercept form that contains point

Nadusha1986 [10]

Slope intercept form is: y=mx+b

For the other line to be parallel they must have the same slope so:

y=5x+b, using the point (4,5) we can solve for the y-intercept, or "b"

5=5(4)+b

5=20+b

b=-15 so the line is:

y=5x-15

5 0

3 years ago