Answer:
The value of
.
Step-by-step explanation:
Diagram of the given scenario shown below.
Given that,
Distance from bench to Jasmyn is
.
Distance from bench to Willard is
.
From the Question,
The given wishing well is a circle and all the distance are tangent to the circle.
So, Triangle formed by these points are such as Δ
and Δ
.
Now, taking both Δ
and Δ
.
⇒
{Radius of circle are equal}
⇒
{Common side}
⇒ ∠
∠
{radius is always perpendicular
to the point of tangent }
∴ Δ
≅ Δ
{By SAS congruence theorem}
Therefore,
....(1) {corresponding part of
congruence triangle (CPCT)}
Thus,
⇒
{from equation (1)}
⇒
⇒ 
Hence, The value of
.
Answer:
max=50
min=41
Step-by-step explanation:
graph
Answer:hmm I don’t fully understand what you mean because there is no bottom of the number line it started at zero so all you have to do is fill the line in.
Step-by-step explanation:
Answer:
yes
Step-by-step explanation:
3x=-9
3x/3 -9/3
x=-3
4x=-12
4x/4 -12/4
x=-3
ps i need brainliest so... plzzz
Answer:
x = 14
y = 6
Step-by-step explanation:
<em>*</em><em>Property</em><em>*</em>
In a parallelogram:
Opposite sides are equal.
Since, the given figure is a parallelogram we can apply this property and equate:
- 1) 2y + 1 = 13
- 2) x = y + 8
<u>1) 2y + 1 = 13</u>
<em>subtracting</em><em> </em><em>1</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>sides</em><em>:</em>
=> 2y + 1 - 1 = 13 - 1
=> 2y = 12
<em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>2</em><em>:</em>
=> 2y/ 2 = 12/ 2
<em>On</em><em> </em><em>the</em><em> </em><em>LHS</em><em>,</em><em> </em><em>2</em><em> </em><em>gets</em><em> </em><em>canceled</em><em> </em><em>because</em><em> </em><em>it's</em><em> </em><em>available</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>numerator</em><em> </em><em>as</em><em> </em><em>well</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>denominator</em><em> </em>
<em>On</em><em> </em><em>the</em><em> </em><em>RHS</em><em>,</em><em> </em><em>2</em><em> </em><em>gets</em><em> </em><em>canceled</em><em> </em><em>when</em><em> </em><em>12</em><em> </em><em>is</em><em> </em><em>factorised</em><em> </em><em>to</em><em> </em><em>6</em><em> </em><em>×</em><em> </em><em>2</em><em>:</em>
<h3>=> y = 6</h3>
<u>2</u><u>)</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>y</u><u> </u><u>+</u><u> </u><u>8</u>
We've got the value of y from equation 1).
<em>Substituting it here:</em>
=> x = 6 + 8
<h3>=> x = 14 </h3>
Answer:
Hence, we got the values of x and y as:
