Can someone help!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

just olya [345]3 years ago

3 0

Answer: A. and D.

Step-by-step explanation: Sorry if its wrong but that sounds right. The first week comes before the second week.

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FIRST CORRECT ANSWER WILL GET BRANLIEST!

il63 [147K]

Greatest common factor (or denominator) is 6,
least common multiple is 180

hmm
(note: I spent like 30 mins trying to use a math only of finding the values but it didn't work so I did a force brute and elimination method explained below)

so
a and b must be multiples of 6
so list all the multiples of 6
wait
180=6*30 and 30's factors are 1,2,3,5,6,10,15,30 so only list the numbers that are the result of multiplying 6 and any of those numbers in that list (so we can have the lcm of 180)
so
6*1=6
6*2=12
6*3=18
6*5=30
6*6=36
6*10=60
6*15=90
6*30=180
these are our possible candidates for the 2 numbers
now we must find the pair that has a GCD of only 6

doing the math is long and tedious so do it yourself (trial and error)

we see that our choices that fulfill both requirements (GCD of 6 and LCM of 180) are
90&12
60&18
30&36
sum them to find the least one

90+12=102
60+18=78
30+36=66

the least possible sum is 66

3 0

3 years ago

A square with an area of 64 in2 is rotated to form a cylinder. What is the radius of the cylinder?

romanna [79]

The area of the square is calculated through the equation,

Area = e²

where e is the measure of sides.

From the given,

64 = e²
e = sqrt 64 = 8 inches

The length of the side of the square rotated to form the cylinder is the diameter of the base of the cylinder. To get the radius, we divide the diameter by 2.

radius = 8 inches / 2 = 4 inches

Thus, the radius of the cylinder is equal to <em>4 inches</em>.

5 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

Stan had 15 apples and gave 10 to his sister. If he had 21 apples and wanted to give the same percentage to his sister, how many

DENIUS [597]

10 x

15 100

15x=1000

x= 66.6666

ANSWER IS 14

8 0

2 years ago

What happens when the multiplicity of a real root is odd?

chubhunter [2.5K]

Answer:

OC. the graph crosses the x-axis

8 0

2 years ago

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