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3 years ago

Complete the statement. Round to the nearest hundredth if necessary.

Mathematics

2 answers:

nika2105 [10]3 years ago

8 0

Answer:

6 cm = 2.36 in (rounded to the nearest hundred)

Step-by-step explanation:

hope this helps :)

umka2103 [35]3 years ago

3 0

Answer:

2.36 inches

Step-by-step explanation:

1. Convert the measurement to inches.

Divide by 2.54

2. 6/2.54= 2.362

3. Look at the number in the thousandths place. If it is larger than 4, round the hundredths up. If it is 4 or smaller, it stays the same.

The number in the thousandths place is a 2, so it stays the same. 2.36

4. 2.36 inches is your answer

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HELP PLS TT^TT 20 POINTS

ki77a [65]

Answer:

Through ASA axiom.

Step-by-step explanation:

It should be by ASA axiom.

The ASA Theorem says: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent.

5 0

3 years ago

Which of the terms cannot be combined with the others? 11 x x -3 x 4 x 2

tiny-mole [99]

Answer: D. 4x^2

Hope this helps!

5 0

3 years ago

Find the value of x so that the polygons have the same perimeter

mestny [16]

x - 2 + x + 4 + x - 2 + 4 - 4 = x + 3 + x + 4 + x + 5

4x + 4 = 3x + 12

4x - 3x = 12 - 4

x = 8

6 0

3 years ago

Find the extreme values of f subject to both constraints f(x,y) = 2x^2+3y^2-4x-5, x^2 + y^2 <=16

Softa [21]

$f(x,y)=2x^2+3y^2-4x-5$
$f_x=4x-4=0\implies x=1$
$f_y=6y=0\implies y=0$

$f(x,y)$ has only one critical point at $(x,y)=(1,0)$ . The function has Hessian

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}4&0\\0&6\end{bmatrix}$

which is positive definite for all $(x,y)$ , which means $f(x,y)$ attains a minimum at the critical point with a value of $f(1,0)=-7$ .

To find the extrema (if any) along the boundary, parameterize it by $x=4\cos t$ and $y=4\sin t$ , with $0\le t$ . On the boundary, we have

$f(x(t),y(t))=F(t)=2(4\cos t)^2+3(4\sin t)^2-4(4\cos t)-5=32\cos^2t+48\sin^2t-16\cos t-5$
$F(t)=35-16\cos t-8\cos2t$

Find the critical points along the boundary:

$F'(t)=16\sin t+16\sin2t=16\sin t+32\sin t\cos t=16\sin t(1+2\cos t)=0$
$\implies t=0,\dfrac{2\pi}3,\pi,\dfrac{4\pi}3$

Respectively, plugging these values into $F(t)$ gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when $F(t)=47$ .

Now, solve for $x,y$ for both cases:

$t=\dfrac{2\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=2\sqrt3\end{cases}$

$t=\dfrac{4\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=-2\sqrt3\end{cases}$

so $f(x,y)$ has two absolute maxima at $(x,y)=(-2,\pm2\sqrt3)$ with the same value of 47.

5 0

3 years ago

Divide 2.548÷52 write an equarion using partial quotients. explain

Llana [10]

19.2 is the answer because if you move the decimal over until its a whole then just divide

6 0

3 years ago