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3 years ago

How many eighth-size parts do you need to model 3/4? explain

1 answer:

6 0

To know the amount of eights you want to get something that looks like ?/8. Basically, the bottom number needs multiplying by 2 (8 divided by 4 is 2). To keep the fraction accurate you multiply by 2 the top and bottom number so you get:
<span>3/4=(3x2)/(4x2)=6/8</span>

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Suppose a line has slope -2/3 and passes through the point (1, 8). which other point must also be on the graph?

Arisa [49]

<h3>Answer: Choice A. (7,4)</h3>

============================================

Explanation:

Use the slope and given point to find the y intercept

y = mx+b

8 = (-2/3)*(1) + b

8 = -2/3 + b

8 + 2/3 = b

24/3 + 2/3 = b

26/3 = b

b = 26/3

The equation of the line is y = (-2/3)x + 26/3

To confirm this, plug in x = 1 and we should get y = 8, due to the point (1,8)

y = (-2/3)x + 26/3

y = (-2/3)*1 + 26/3

y = -2/3 + 26/3

y = (-2+26)/3

y = 24/3

y = 8

So that verifies we have the correct equation.

--------------

Next, go through each answer choice to see if the x coordinate of the point leads to the y coordinate.

If we try x = 7, then,

y = (-2/3)x + 26/3

y = (-2/3)(7) + 26/3

y = -14/3 + 26/3

y = (-14+26)/3

y = 12/3

y = 4

This shows that (7,4) is on the line. Choice A is the answer

That rules out choice B.

----------------

If we tried x = -5, then,

y = (-2/3)x + 26/3

y = (-2/3)(-5) + 26/3

y = 10/3 + 26/3

y = 36/3

y = 12

meaning that (-5,12) is on the line. That rules out choices C and D.

Refer to the graph below. It visually confirms that of the four answer choices, only point A is on the line. I used GeoGebra to make the graph.

5 0

3 years ago

Which sequences of transformations will map circle<br><br> M onto circle N?

sergiy2304 [10]

Answer:

1: Reflect M across the x-axis

2: Dilate about the center by 3/2

Step-by-step explanation:

Given

See attachment for M and N

Required

Which maps M to N

The coordinates of the radius of the circles are:

$M = (5,5)$

$N = (5,-5)$

And the radius of circles are:

$r_M=2$

$r_N=3$

The first transformation from M to M' is:

- Reflect across the x-axis

The rule is:

$(x,y) \to (x,-y)$

$M(5,5) \to M'(5,-5)$

<em>At this point, M' and N now have the same center but different radius.</em>

The second transformation from M' to N is:

- Dilate about the center by dividing the radius of N by the radius of M

i.e.

$k =\frac{r_N}{r_M}$

$k =\frac{3}{2}$

<em>At this point, M has been completely mapped to N.</em>

4 0

3 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

3 years ago

Look at pic plz help

jasenka [17]

Answer:

Mochi

Step-by-step explanation:

Mochi

Each week theres a increase of 4 in Mochi's weight. So you will add 4 twice to get to the 5th week.

4th week : 17

5th week : 21

Kappa

In 4 weeks Kappa weighed 14, which is less that Mochi at the pace Kappa is going at.

on the 3rd week Kappa weighed 11

and 2nd week Kappa weighed 5

and 1st week Kappa weighed 3

Looking at the graph, we can give a rough estimate on the 5th week it's going to be 16 or 17 either way making Mochi heavier.

8 0

3 years ago

Please solve this problem tnx....

True [87]

Answer:

27.50

Step-by-step explanation:

Since it is 16.50 each and they bought 5 packets total an easy way to solve this problem would be that you multiply 16.50 × 5 which will give you how much the total cost of the candy would be:

16.50 × 5 = 82.50

Since that is the total amount you would then divide 82.50 by 3 since Mary and her friends all split the cost three ways (since there are three of them):

82.50 ÷ 3 = 27.50

Which will give you your answer.

7 0

3 years ago