Answer:
DG = 30
Step-by-step explanation:
Given:
DH = 6
DE = 4
EF = 16
Required:
DG
Solution:
DG = DH + HG
DG = 6 + HG
Let's find HG
Given that HE is parallel to the third side of ∆DGF, based on the side-splitter theorem, the other two sides of ∆DGF are divided proportionally.
Therefore,
DH/HG = DE/EF
6/HG = 4/16
Cross multiply
HG*4 = 16*6
HG = 96/4
HG = 24
✔️DG = 6 + HG
DG = 6 + 24
DG = 30
Answer:
5 times 2
Step-by-step explanation:
5 times 2 because I know my multiplication. Even if I didn't know my ti.es table I can just look at it
Answer: (a)
(b) 
Step-by-step explanation:
(a) P( Bill hitting the target) = 0.7 P( Bill not hitting the target) = 0.3
P( George hitting the target) = 0.4 P(George not hitting the target) = 0.6
Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3
= 0.54
Chances that George hit the target is = 0.4 x 0.3 = 0.12
So given that exactly one shot hit the target, probability that it was George's shot =
=
.
(b) The numerator in the second part would be the same as of (a) part which is 0.12.
The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.
Given that the target is hit,probability that George hit it =
= =
Answer:
5x+8y
Step-by-step explanation:
We can first open the brackets by using the distrubitive propety.
2(x+4y)+3x
2x+8y+3x
Now we can combine like terms.
5x+8y.
I combined the 3x and the 2x. This works because imagined I have 3 of something then I got 2 more of that something I would now have 5 of that something.
Hoped this helped,
JoeLouis2