We see from the attached, that kite area = product of the diagonals / 2
The diagonals could be 12 by 8 or
6 by 16 or
3 by 32, etc
It cannot be narrowed down any further.
Answer:
14 and 4/1000 or 14004/1000
Step-by-step explanation:
1*10+3*1+4*(1/1000)
First i like to separate the equation by PEMDAS or order of operations. First is parentheses, which is the fraction. Then is exponents. There is none. Then multiplication and division so u solve 1*10, 3*1 and 4*1/1000.
This is the updated equation: 10+4+(4/1000)
Next you do subtraction and addition. The order does not matter if there's only addition. If there is subtraction u solve by the order of the equation. SO in this step we solve 10+4+(4/1000)
Here is the updated version: 14+(4/1000)
Then u add that which is 14 and 4/1000 or you can do 14004/1000
Answer:
C. The 6th term is positive/negative 80
Step-by-step explanation:
Given
Geometric Progression
Required
To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;
To solve the common ratio;
Divide the 7th term by the 5th term; This gives
Divide the numerator and the denominator of the fraction by 40
----- equation 1
Recall that the formula of a GP is
Where n is the nth term
So,
Substitute the above expression in equation 1
becomes
Square root both sides
r = ±
Next, is to solve for the first term;
Using
By substituting 160 for T5 and ± for r;
We get
Multiply through by 16
Now, we can easily solve for the 6th term
Recall that the formula of a GP is
Here, n = 6;
r = ±
So,
or 
or 
or 
±80
Hence, the 6th term is positive/negative 80
Answer:
D.
Each line has the 3 dashes meaning equivalent
<h3>
Answer:</h3>
∠BDC and ∠AED are right angles
<h3>
Step-by-step explanation:</h3>
Because ∠C ≅ ∠C, the additional bit of information above can be used to show AA similarity.
____
None of the other offered choices says anything about <em>both</em> triangles. In order to show similarity, you need information about corresponding parts of the <em>two</em> triangles. Information about one triangle alone is not sufficient.
