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3 years ago

Determine whether AB is tangent to the circle with center c

Mathematics

1 answer:

GenaCL600 [577]3 years ago

3 0

9514 1404 393

Answer:

yes, no, no

Step-by-step explanation:

The triangle will be a right triangle if the sum of the squares of the two short sides, less the square of the long side, is zero. (This is what the Pythagorean theorem tells you.) The angle where the radius meets the circle must be a right angle if the line is to be a tangent.

a) 3^2 +4^2 -5^2 = 9 + 16 -25 = 0 . . . . AB is a tangent

b) 9^2 +15^2 -18^2 = 81 +225 -324 = -18 . . . . AB is not a tangent

c) 10^2 +48^2 -52^2 = 100 +2304 -2704 = -300 . . . . AB is not a tangent

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Is -91.51 an integer

Anna71 [15]

No it’s not an integer i looked it up on google.

6 0

4 years ago

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What is the value of the arithmetic series below?

xxTIMURxx [149]

Answer:

(A)-494

Step-by-step explanation:

Given the arithmetic series

$S_{19}=\sum_{k=1}^{19}4-3k$

The terms in the sequence are:

When k=1, 4-3k=4-3(1)=1
When k=2, 4-3k=4-3(2)=-2
When k=3, 4-3k=4-3(3)=-5

Therefore, the terms in the sequence are: 1, -2, -5, ...

First term, a =1

Common difference, d=-2-1=-3

The sum of an arithmetic series, $S_n=\dfrac{n}{2}[2a+(n-1)d]$

Therefore:

$S_{19}=\dfrac{19}{2}[2(1)+(19-1)(-3)]\\=9.5[2+18*-3]\\=9.5[2-54]\\=9.5*-52\\=-494$

The correct option is A.

6 0

3 years ago

Two thirds of a number decreased by six is two. what is the number?

Kisachek [45]

Answer: The number is: " 12 ".

____________________________________
Let "x" represent "the unknown number" (for which we wish to solve.

The expression:

$\frac{2}{3}$ x − 6 = 2 ; Solve for "x" ;
_______________________________________________
Method 1)

Add "6" to EACH SIDE of the equation;
_______________________________________________
→ $\frac{2}{3}$ x − 6 + 6 = 2 + 6 ;

to get:

→ $\frac{2}{3}$ x = 8 ;
______________________________________________
Multiply each side of the equation by " $\frac{3}{2}$ " ; to isolate "x" on one side of the equation ; and to solve for "x" ;
______________________________________________
→ $\frac{3}{2}$ * $\frac{2}{3}$ x = 8 * $\frac{3}{2}$ ;

→ x = 8 * $\frac{3}{2}$ ;

= $\frac{8}{1}$ * $\frac{3}{2}$ ;

= $\frac{8*3}{1*2}$ ;

= $\frac{24}{2}$ ;

= 12 .
______________________________________________
x = 12 .
______________________________________________
Method 2)
______________________________________________
$\frac{2}{3}$ x − 6 = 2 ; Solve for "x" ;

Add "6" to EACH SIDE of the equation;
_______________________________________________
→ $\frac{2}{3}$ x − 6 + 6 = 2 + 6 ;

to get:
→ $\frac{2}{3}$ x = 8 ;
______________________________________________
Multiply each side of the equation by "3" ; to get rid of the "fraction" ;
 → 3 * $\frac{2}{3}$ x = 8 * 3 ;
→ $\frac{3}{1}$ * $\frac{2}{3}$ x = 8 * 3 ;
→ $\frac{3*2}{1*3}$ x = 8 * 3
→ $\frac{6}{3}$ x = 24 ;

→ 2x = 24 ;

→ Divide each side of the equation by "2" ; to isolate "x" on one side of the equation; & to solve for "x" :

2x / 2 = 24 / 2 ;

x = 12 .
__________________________________________________
Method 3).
__________________________________________________
$\frac{2}{3}$ x − 6 = 2 ; Solve for "x" ;
_______________________________________________
Add "6" to EACH SIDE of the equation;
_______________________________________________
→ $\frac{2}{3}$ x − 6 + 6 = 2 + 6 ;

to get:

→ $\frac{2}{3}$ x = 8 ;
______________________________________________
Now, divide each side of the equation by " $\frac{2}{3}$ " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
___________________________________________________
{ $\frac{2}{3}$ x } / { $\frac{2}{3}$ } = 8 / { $\frac{2}{3}$ } ;

to get: x = 8 / { $\frac{2}{3}$ } ;

= 8 * ( $\frac{3}{2}$ ;

= $\frac{8}{1}$ * $\frac{3}{2}$ ;

= $\frac{8*3}{1*2}$ ;

= $\frac{24}{2}$ ;

= 12 ;
___________________________________________
x = 12 .
___________________________________________
NOTE: Variant: (in "Methods 2 & 3") :
___________________________________________
At the point where:
___________________________________________
= 8 * ( $\frac{3}{2}$ ) ;

= $\frac{8}{1}$ * $\frac{3}{2}$ ;
__________________________________________
We can cancel out the "2" to a "1" ; and we can cancel out the "8" to a "4" ;
__________________________________________
{since: "8÷2 = 4" ; and since: "2÷2 =1" } ;
__________________________________________
and we can rewrite the expression:
__________________________________________
$\frac{8}{1}$ * $\frac{3}{2}$ ;
__________________________________________
as: $\frac{4}{1}$ * $\frac{3}{1}$ ;
__________________________________________
which equals:
__________________________________________
→ $\frac{4*3}{1*1}$ ;

= $\frac{12}{1}$ ;

= 12 .
__________________________________________
x = 12 .
__________________________________________
Answer: The number is: " 12 ".
__________________________________________

8 0

4 years ago

What are the perpendicular sides?

inn [45]

Answer:

Step-by-step explanation: they are lines that form a right angle

8 0

3 years ago

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The ratio of girls to boys in the ninth grade is 5:4. if there are 144 boys in the ninth grade, how many girls are there?

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