..............................

Estimate sum 542+328=

Ghella [55]

542 --> 500

328 --> 300

Estimate: 800

8 0

3 years ago

Concept maps work because they _______.

AnnZ [28]

Showing the relationships among concepts.

7 0

3 years ago

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Four friends ordered two pizzas the cost of each pizza is $13.80 about how much will it cost for both pizzas

skad [1K]

13.80 x 2= 27.60

1 pizza equals 13.80 so two pizzas will be twice that money amount. Both pizzas will cost $27.60.

I hope this helped

8 0

3 years ago

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Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

Marina86 [1]

Answer:

The probability of selecting a black card or a 6 = 7/13

Step-by-step explanation:

In this question we have given two events. When two events can not occur at the same time,it is known as mutually exclusive event.

According to the question we need to find out the probability of black card or 6. So we can write it as:

P(black card or 6):

The probability of selecting a black card = 26/52

The probability of selecting a 6 = 4/52

And the probability of selecting both = 2/52.

So we will apply the formula of compound probability:

P(black card or 6)=P(black card)+P(6)-P(black card and 6)

Now substitute the values:

P(black card or 6)= 26/52+4/52-2/52

P(black card or 6)=26+4-2/52

P(black card or 6)=30-2/52

P(black card or 6)=28/52

P(black card or 6)=7/13.

Hence the probability of selecting a black card or a 6 = 7/13 ....

7 0

3 years ago

A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be

irga5000 [103]

Answer:

The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

Step-by-step explanation:

Let $C = (h,k)$ the coordinates of the center of the circle, which must be transformed into $C'=(h', k')$ by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

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

Where $r$ is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) Center of the circle is translated 4 units to the right (+x direction):

$C''(x,y) = C(x, y) + U(x,y)$ (Eq. 1)

Where $U(x,y)$ is the translation vector, dimensionless.

If we know that $C(x, y) = (h,k)$ and $U(x,y) = (4, 0)$ , then:

$C''(x,y) = (h,k)+(4,0)$

$C''(x,y) =(h+4,k)$

2) Reflection over the x-axis:

$C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)]$ (Eq. 2)

Where $O(x,y)$ is the reflection point, dimensionless.

If we know that $O(x,y) = (h+4,0)$ and $C''(x,y) =(h+4,k)$ , the new point is:

$C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]$

$C'(x,y) = (h+4, 0)-(0,k)$

$C'(x,y) = (h+4, -k)$

And thus, $h' = h+4$ and $k' = -k$ . The statement is now presented as:

$\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}$

In other words, this mathematical statement can be translated as:

There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r.

4 0

3 years ago