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3 years ago

1. 7/8 + 1 1/2 = 2. 268.97 + 38.659 = 3. 2 3/4 - 2/3 = 4. 870.45- 96.38=

Mathematics

2 answers:

Mariana [72]3 years ago

8 0

2. Equals 307.629 because if you line up the decimals when you and then you should get your right answer

MissTica3 years ago

5 0

I boxed all of the answers. sorry if it’s hard to read but yeah

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If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 930

Free_Kalibri [48]

Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.

Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
$s=ut+ \frac{1}{2} at^2$

But, we assume that the train accelerates from rest, thus
$s=\frac{1}{2} at^2 \\ \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\ \\ \Rightarrow t^2=534.48 \\ \\ \Rightarrow t=\sqrt{534.48}=23.12$

The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).

Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.

6 0

3 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

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

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

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

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

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

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

Please hurry i don't have much time left!

quester [9]

Answer:

Check below

Step-by-step explanation:

That's too bad you haven't attached a rectangle.

Here's an example, with the data you've typed in.

1) When we dilate a rectangle we either grows it or shrink it through a scale factor.

Check the first picture below.

The New Dilated Rectangle A'B'C'D' will follow its coordinates, when the <u>Center of Dilation is at its origin(Middle):</u>

$D_{A'B'C'D'}=\frac{1}{2}(x,y)$

2) But In this question, <u>B is the center of Dilation</u>. So, Since B is the Dilation Point B=B' . And More importantly:

$\bar{AB}=\frac{1}{2}\bar{A'B'}\\\bar{CD}=\frac{1}{2}\bar{C'D'}\\\bar{AC}=\frac{1}{2}\bar{A'C'}\\\bar{CD}=\frac{1}{2}\bar{C'D'}\\$

3) So check the pictures below for a better understanding.

6 0

3 years ago

Read 2 more answers

I think number 9 is b or c, i don't know what 10 is. please help me!!!

scoray [572]

10 is letter b. I hope this helps!!!

7 0

3 years ago

Solve the exponential question . Leave your answer as a fraction

Doss [256]

Step-by-step explanation:

problem → 2^x = 4, solve for x

⇒2^x=4

⇒2^x=2^2

⇒x=2

7 0

3 years ago