ACB, ACD and DCB are all right triangles. We can then represent the relationships of their legs through the Pythagorean theorem.
For triangle ACB:
7^2=(AC)^2+(BC)^2 (1)
For triangle ACD:
(AC)^2=3^2+(CD)^2 (2)
For triangle DCB:
(BC)^2=(CD)^2+4^2 (3)
To make things simpler, we let
(AC)=x
(BC)=y
(CD)=z
Hence we have:
49=x^2+y^2 (1)
y^2=z^2+16 (2)
x^2=9+z^2 (3)
To solve for the unknown c, let us first substitute equation (2) in (1)
49=x^2+z^2+16
33=x^2+z^2 (4)
Then, we can substitute equation (3) in (4)
33=9+z^2+z^2
24=2z^2
z^2=12
z=2sqrt(3)
Thus, the length of z or (CD) is 2sqrt(3) in simplest radical form.
No, because 5 subtracted from 7 is 4. 7-5=4
Answer:
f(-1) = 11
Step-by-step explanation:
f(x) = 2x^2 - 6x + 3
Let x = -1
f(-1) = 2(-1)^2 -6(-1) +3
=2*1 +6 +3
=2+6+3
= 11
In the first attachment your answer is A.
In the second attachment your answer is D.
In the third attachment your answer is D.
In the fourth attachment your answer is 90cm
In the fifth attachment your answer is 98<span />
The correct answer is =7y^2
hope i helped^-^