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3 years ago

9.6 dm= _____ cm

Mathematics

2 answers:

Sergio039 [100]3 years ago

8 0

Answer:

96 cm

53 cm

4,5 Cm

1,5 Cm

1.7dm

0,5m

Step-by-step explanation:

Dm to Cm ×10

mm to cm ÷10

Sever21 [200]3 years ago

5 0

93 cm
53 cm
4,5 cm
1,5 cm
1.7 dm
0,5 m

Hope this helped :)

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Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation:

$r^2=\Delta x^2+\Delta y^2$

In this case, we will solve the delta with the center coordinate and the B coordinate.

$\begin{gathered} r^2=((-4)-(-8))^2+((-10)-(-7)) \\ r^2=(-4+8)^2+(-10+7)^2 \\ r^2=4^2+(-3)^2 \\ r^2=16+9 \\ r^2=25 \\ r=5 \end{gathered}$

Therefore, the equation for the standard form of a circle is:

$\begin{gathered} (x-(-8))^2+(y-(-7))^2=25 \\ (x+8)^2+(y+7)^2=25 \end{gathered}$

In conclusion, the equation is the following:

$(x+8)^2+(y+7)^2=25$

4 0

1 year ago

Help please and thank you

marysya [2.9K]

Answer:

Step-by-step explanation:

4/5

1/2

8 0

3 years ago

Read 2 more answers

Evaluate the integral ∫2−1|x−1|dx

defon

I think you might be referring to the definite integral,

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx$

Recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

Then $|x-1|=x-1$ if $x\ge1$ , and $|x-1|=1-x$ is $x$ . So spliting up the integral at <em>x</em> = 1, we have

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx$

The rest is simple:

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}$

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4x – 7y + 3z - 10<br><br> What's the coefficient of the 2nd term?

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