Answer is B. second one
A = 1/8, B = 5/16, C = 7/12, D = 5/6
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Answer:
<em>f(x)=x²-3x-10</em>
Step-by-step explanation:
\begin{gathered}f(x) = x {}^{2} - 3x - 10 \\ to \: find \: x \: intercept \:o r \: zero \: substitute \: f(x) = 0\: \\ 0 = x {}^{2} - 3x - 10 \\ x {}^{2} - 3x - 10 = 0 \\ x {}^{2} + 2x - 5x - 10 = 0 \\ x(x + 2) - 5x - 10 = 0 \\ x(x + 2) - 5(x + 2) = 0 \\ (x + 2).(x - 5) = 0 \\ x + 2 = 0 \\ x - 5 = 0 \\ x = - 2 \\ x = 5\end{gathered}
f(x)=x
2
−3x−10
tofindxinterceptorzerosubstitutef(x)=0
0=x
2
−3x−10
x
2
−3x−10=0
x
2
+2x−5x−10=0
x(x+2)−5x−10=0
x(x+2)−5(x+2)=0
(x+2).(x−5)=0
x+2=0
x−5=0
x=−2
x=5
therefore the zeros of the equation are x₁=-2,x₂=5
Answer:
The answer is 2 1/6
Step-by-step explanation:
First you would divide 13 by six and that would leave you with 2 with a remainder of 1. Then you would put together the fraction leaving 6 as the denominator, 2 as the whole number, and 1 as the numerator.
Let me know if you need any other help!
Answer:
C
Step-by-step explanation:
To make it easy let's start by organizing our information :
- AC=12 AND BD=8
- ABCD is a rhombus
- K and L are the midpoints of sides AD and CD
- we notice that the rhombus ABCD is divided into four right triangles
What do you think of when you hear a right triangle ?
- The pythagorian theorem !
AC and BD are khown so let's focus on them .
If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?
Simply because they are the diagonals of a rhombus .
ow let's apply the pythagorian theorem :
- (AC/2)² + (BD/2)² = BC²
- 6²+4²=52
- BC²= 52⇒
=BC
Now we khow that : AB=BC=CD=AD=
This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle
- KL is parallel to AC
- k is the midpoint of AD and L of DC
I smell something . yes! Thales theorem
- KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
- KL/12=
/2*
- KL/12=1/2⇒ KL=6
Now we have the length of the base kl
Now the big boss the height :
- notice that you khow the length of KL
- BD crosses kl from its midpoint and DL =
/2
What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D
- DL²=(KL/2)²+D²
- 52/4= 9+ D²
- D² = 52/4-9 +4 SO D=2
now the height of the trigle is H= BD-D= 8-2=6
NOw the area of the triangle is :
- A=(KL*H)/2 ⇒ A= (6*6)/2=18
THE ANSWER IS 18 SQ.UN