Identify all the drawn radii of the circle created by the diameter and is perpendicular bisector

Answers

Answer 1

Answer:

It’s a personal challenge

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-\dfrac{1}{6} \times \left(-\dfrac{9}{7}\right)− 6 1 ×(− 7 9 )minus, start fraction, 1, divided by, 6, end fraction, times, left parenthesis, minus, start fraction, 9, divided by, 7, end fraction, right parenthesis

Answers

Answer:

$(3)/(14)$

Step-by-step explanation:

The even number of minus signs means the product will be positive. A factor of 3 can be removed to simplify the product. As usual, the numerator of the product is the product of numerators, and the denominator of the product is the product of denominators.

$-(1)/(6) * \left(-(9)/(7)\right)=(1\cdot 9)/(6\cdot 7)=(3\cdot 3)/(3\cdot 14)=\boxed{(3)/(14)}$

The answer to the expression $-(1)/(6) * \left(-(9)/(7)\right) - (1)/(6) * \left(-(9)/(7)\right)$ is 0.

In this expression, we have two terms that are identical: $-(1)/(6) * \left(-(9)/(7)\right)\) and \(-(1)/(6) * \left(-(9)/(7)\right)$ .

When we multiply a negative number by another negative number, the result is positive.

So, both of these terms become positive, and since they are identical, when we subtract one from the other, we get zero.

Now, let's provide a more detailed explanation.

When we multiply $-(1)/(6)\) by \(-(9)/(7)$ , we first multiply the numerators to get 9 and the denominators to get 42.

So, $-(1)/(6) * \left(-(9)/(7)\right) = (9)/(42)$ .

Now, we can simplify $(9)/(42)$ by dividing both the numerator and denominator by their greatest common divisor, which is 3.

So, $(9)/(42)$ simplifies to $(3)/(14)$ .

Therefore, the expression becomes $(3)/(14) - (3)/(14)$ , and when we subtract these two identical fractions, the result is indeed 0.

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Need the values IN FRACTIONS (no mixed numbers or decimals)1. -1/2^1
2. -1/2^2
3. -1/2^3
4. -1/2^4
5. -1/2^5

Answers

$\left(-(1)/(2)\right)^1=-(1)/(2)\n \left(-(1)/(2)\right)^2=(1)/(4)\n \left(-(1)/(2)\right)^3=-(1)/(8)\n \left(-(1)/(2)\right)^4=(1)/(16)\n \left(-(1)/(2)\right)^5=-(1)/(32)\n$

If (-10) is divided by (-2), what is the quotient? a) 2 b) -2 c) 5 d) -5

Answers

Answer:5

Step-by-step explanation:

we know two negatives divided by each other make a positive so -10/-2 is the equivalent of 10/2= 5

Hank and his two friends are attending a concert. They purchase tickets and parking for a total of 129.00 they decide to split it evens my how much does each person owe

Answers

Answer:

$43.00

Step-by-step explanation:

$129.00 / 3=$43.00

Given the equation 4x + 2y = 9, which equation below would cause an inconsistent-independent system?A:6x + 12y = 13
B: 2x + y = -6
C:x - 2y = 7
D:8x - 4y = -5

Answers

Inconsistent-independent system means that there are no solutions.
Graphically the lines of those equations will never cross - the lines are parallel.

So when the two lines are parallel?
When they have the same A factors.

So when you have an equation 4x+2y=9 we need to rewrite it to
y= -2x+9/2
So the A factor is -2 (it is the number with x) and the B factor is 9/2

So all we need to do now is to find an equation with the same A factor.

Lets rewrite equations in answers:
A: 6x+12y=13 -> y=-1/2x+13/12 A=-1/2 - so it is not an equation we are looking for
B: 2x+y=-6 -> y=-2x-6 A=-2 - this is an equation we were looking for.
C: x-2y=7 -> y=1/2x-7/2 A=1/2 - so it is also not a good answer

The answer is B: the equation 2x+y=-6 will cause the inconsistent-independent system with the given equation.

So a 4 digit number is represented ABCD, they are not multiplied together, each letter represents 1 single , whole numberif ABCD times D=DCBA

solve for the values of A and B and C and D with full workings and solution (this problem is from my 5th grade math book, just for fun)

Answers

$\overline{ABCD}=10^3A+10^2B+10C+D\n\n\overline{ABCD}* D=(10^3A+10^2B+10C+D)* D\n=10^3AD+10^2BD+10CD+D^2\n\n\overline{DCBA}=10^3D+10^2C+10B+A\n\n\overline{ABCD}* D=\overline{DCBA}\n\downarrow\n10^3AD+10^2BD+10CD+D^2=10^3D+10^2C+10B+A$

$one\ of\ the\ solutions\ of\ general:\n\n\#1\ AD=D\n\#2\ BD=C\n\#3\ CD=B\n\#4\ D^2=A\n\nfrom\ \#1\ and\ \#4\nAD=D\ and\ D^2=A\to\ conclusion:A=D=1\n\nfrom\ \#2\ and\ \#3\nBD=C\ and\ CD=B\to conclusion:B=C\n\ntherefore$

$\overline{ABCD}=\n1001\n1111\n1221\n1331\n1441\n1551\n1661\n1771\n1881\n1991$

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