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Katena32 [7]
3 years ago
14

The opposite of 1.5 is ? So the opposite 1.5 ? Which is equal to the opposite of ? Which is ? The second statement is ?

Mathematics
1 answer:
rewona [7]3 years ago
7 0
I think the opposite of 1.5 is -1.5
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A carpenter worked 10 weeks on a particular job, 51/2 days per week and 7 3/4 hours per day. How many hours did the carpenter wo
Luden [163]

Answer:

= 426 1/4 hr.

Step-by-step explanation:

5 1/2 = 5.5 Days in a week

7 3/4 = 7.75 hr. Per Day

Total hours in a week = 5.5 × 7.75 = 42.625 hr in a week

In 10 weeks = 42.625 × 10 = 426.25 hr.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20-%205%20%5Csqrt%7B49%20-%208a%20%3D%20%20-%2035%7D%20" id="TexFormula1" title=" - 5 \sqrt{4
Tcecarenko [31]

- 5 \sqrt{49 - 8a =  - 35}  \\  \\ 1. \:  - 8a = 0 \\  \\ 2. \:  \frac{ - 8a}{ - 8} =  \frac{0}{ - 8}   \\ a = 0

7 0
3 years ago
What is the primary goal and secondary goal of the Fried Liver Opening in chess? &lt;br /&gt;
polet [3.4K]

Play usually continues 7.Qf3+ Ke6 8.Nc3 (see diagram). Black will play 8...Nb4 or 8...Ne7 and follow up with c6, bolstering his pinned knight on d5. If Black plays 8...Nb4, White can force the b4 knight to abandon protection of the d5 knight with 9.a3?! Nxc2+ 10.Kd1 Nxa1 11.Nxd5, sacrificing a rook, but current analysis suggests that the alternatives 9.Qe4, 9.Bb3 and 9.O-O are stronger. White has a strong attack, but it has not been proven yet to be decisive.

Because defence is harder to play than attack in this variation when given short time limits, the Fried Liver is dangerous for Black in over-the-board play, if using a short time control. It is also especially effective against weaker players who may not be able to find the correct defences. Sometimes Black invites White to play the Fried Liver Attack in correspondence chess or in over-the-board games with longer time limits (or no time limit), as the relaxed pace affords Black a better opportunity to refute the White sacrifice.


3 0
3 years ago
Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w
siniylev [52]

Solution :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.

$r(t)=x(t)i+y(t)j$  is a parametrization of the most $\text{rapid increase curve}$ , then

$\frac{dx}{dt}=\frac{dx}{dt}i+\frac{dy}{dt}j$ is a tangent to the curve.

So then we have that $\frac{dr}{dt}=\lambda \triangledown C$

$\Rightarrow \frac{dx}{dt}=-\frac{2\lambda x}{10^4}e^{-(x^2+2y^2)/10^4}, \frac{dy}{dt}=-\frac{4\lambda y}{10^4}e^{-(x^2+2y^2)/10^4} $

∴ $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2y}{x}$

Using separation of variables,

$\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int \frac{dx}{x}$

$\ln y=2 \ln x$

⇒ y = kx^2 for some constant k

but we know that $y(x_0)=y_0$

$\Rightarrow kx_0^2=y_0$

$\Rightarrow k =\frac{y_0}{x_0^2}$

∴ The path of the shark will follow is along the parabola

$y=\frac{y_0}{x_0^2}x^2$

$y=y_0\left(\frac{x}{x_0}\right)^2$

7 0
3 years ago
If sin theta = -0.7660, which of the following represents an approximate value of tan theta, for 180 degrees &lt; theta &lt; 270
Elis [28]
We have that
sin ∅=-0.7660

the sin ∅ is negative
so 
∅ belong to the III or IV quadrant
but
180°< ∅ < 270°
hence
∅ belong to the III quadrant
sin ∅=-0.7660
sin² ∅+cos² ∅=1
cos² ∅=1-sin² ∅------> cos² ∅=1-(0.7660)²-----> 0.4132
cos ∅=√0.4132-----> cos ∅=0.6428
the value of cos ∅ is negative-------> III quadrant
cos ∅=-0.6428

tan ∅=sin ∅/ cos ∅----> tan ∅=-0.7660/-0.6428----> tan ∅=1.1917

the answer is
tan ∅ is 1.1917
4 0
3 years ago
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