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dsp73
4 years ago
5

1/4 rounded to the nearest whole number

Mathematics
2 answers:
jek_recluse [69]4 years ago
3 0
The next bigger whole number is 1. The next smaller whole number is 0. 1/4 is closer to 0 than it is to 1. So the nearest whole number is zero.
Fed [463]4 years ago
3 0
Hell there.

<span>1/4 rounded to the nearest whole number

Answer: 0</span>
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Three subtracted from the product of
ololo11 [35]
5w-3 > -29
Add 3 at both sides

5w > -26
Divide 5 at both sides

w > -26/5

6 0
4 years ago
Find the missing value ​
lapo4ka [179]

Answer:

-10

Step-by-step explanation:

since the answer is positive, we need a negative number, if we use -10,two negatives makes a positive, so it becomes

-8 +10=2

4 0
4 years ago
Find the area of a regular hexagon with the given measurement.
yawa3891 [41]

Answer:  24\sqrt{3}in^2 or  41.56in^2

Step-by-step explanation:

You can find the area of a regular hexagon with the following formula:

A_{(hexagon)}=\frac{3\sqrt{3}s^2}{2}

Where "s" is any side of the regular hexagon.

For this hexagon you know that the length of each side is 4 inches. Then, you must substitute s=4in into the formula  A_{(hexagon)}=\frac{3\sqrt{3}s^2}{2}.

Therefore, the area of this regular hexagon is:

 A_{(hexagon)}=\frac{3\sqrt{3}(4in)^2}{2}

 A_{(hexagon)}=24\sqrt{3}in^2 or A_{(hexagon)}=41.56in^2

8 0
3 years ago
Read 2 more answers
Let the (x; y) coordinates represent locations on the ground. The height h of
grigory [225]

The critical points of <em>h(x,y)</em> occur wherever its partial derivatives h_x and h_y vanish simultaneously. We have

h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5

Substitute <em>y</em> in the second equation and solve for <em>x</em>, then for <em>y</em> :

2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}

This is to say there are two critical points,

(x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)

To classify these critical points, we carry out the second partial derivative test. <em>h(x,y)</em> has Hessian

H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}

whose determinant is 192y-16. Now,

• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and h_{xx} are positive at the point, then it's a local minimum

• if the determinant is positive and h_{xx} is negative, then it's a local maximum

• otherwise the test fails

We have

\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0

while

\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0

So, we end up with

h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}

7 0
3 years ago
an earthquake with a magnitude of about 2.0 or less is called a microearthquake. it is not usually felt. the intensity of an ear
Artemon [7]

Answer:

100 times greater

Step-by-step explanation:

i promise

6 0
3 years ago
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