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2 years ago

If the vehicles are divided evenly between the sections, how many vehicles are in each section?

Mathematics

1 answer:

solmaris [256]2 years ago

5 0

That would depend on how many vehicles there are. The question is not complete.

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(-8)x...x(-8) times 20

Stella [2.4K]

Answer:

1280

Step-by-step explanation:

rewrite (-8) times (-8) as an equivalent addition problem which is 8 times 8 = 64

take 64 then times it by 20 which is 64 times 20 equals 1,280

Hope this helps:)

6 0

2 years ago

Find the value of r .

Sever21 [200]

Answer:

0.5

Step-by-step explanation:

hypotenuse/adjacent=r

7/14=0.5

3 0

3 years ago

Visitors to the schools website can follow a link to the schools chorale pictures.If the website had 250 visitors in one week an

bixtya [17]

The answer is 3/10 times 250 which is 75.

3 0

3 years ago

Georgia lifts free weights. The bar weighs 25 pounds. She puts two 25-pound weights and two 10-pound weights on the bar. How muc

zubka84 [21]

First you will add 25+25=50-pounds
Second you will add 10+10=20-pounds
Third you will add 50+20=70-pounds for all the weights
Fourth you will add the total pounds of all the weights and the weight of the bar 70+25=95
So, the answer would be 95-pounds.
Hope this helps! Have a good day!

7 0

2 years ago

Read 2 more answers

show that the equation 3x2 + 3y2 + 6x-y= 0 represents a circle and find the centre and radius of the circle

Jlenok [28]

Answer:

Center of the circle: $(-1,\frac{1}{6})$

Radius of the circle: $\frac{\sqrt{37} }{6}$

Step-by-step explanation:

Let's start by dividing both sides of the equation by the factor "3" so we simplify our next step of completing squares for x and for y:

$3x^2+3y^2+6x-y=0\\x^2+y^2+2x-\frac{1}{3} y=0$

Now we work on completing the squares for the expression on x and for the expression on y separately, so we group together the terms in "x" and then the terms in "y":

$x^2+y^2+2x-\frac{1}{3} y=0\\( x^2+2x ) + (y^2-\frac{1}{3} y)=0$

Let's find what number we need to add to both sides of the equation to complete the square of the group on the variable "x":

$( x^2+2x ) = 0\\x^2+2x+1=1\\(x+1)^2=1$

So, we need to add "1" to both sides in order to complete the square in "x".

Now let's work on a similar fashion to find what number we need to add on both sides to complete the square for the group on y":

$(y^2-\frac{1}{3} y)=0\\y^2-\frac{1}{3} y+\frac{1}{36} =\frac{1}{36}\\(y-\frac{1}{6} )^2=\frac{1}{36}$

Therefore, we need to add " $\frac{1}{36}$ " to both sides to complete the square for the y-variable.

This means we need to add a total of $1 + \frac{1}{36} = \frac{37}{36}$ to both sides of the initial equation in order to complete the square for both variables:

$x^2+y^2+2x-\frac{1}{3} y=0\\x^2+y^2+2x-\frac{1}{3} y+\frac{37}{36} =\frac{37}{36} \\(x+1)^2+(y-\frac{1}{6} )^2=\frac{37}{36}$

Now recall that the right hand side of this expression for the equation of a circle contains the square of the circle's radius, based on the general form for the equation of a circle of center $(x_0,y_0)$ and radius R:

$(x-x_0)^2+(y-y_0)^2=R^2$

So our equation that can be written as:

$(x+1)^2+(y-\frac{1}{6} )^2=(\sqrt{\frac{37}{36}} )^2$

corresponds to a circle centered at $(-1,\frac{1}{6})$ , and with radius $\sqrt{\frac{37}{36}}=\frac{\sqrt{37} }{6}$

5 0

2 years ago