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likoan [24]
2 years ago
10

Which congruence postulate is this? SAS, SSS, AAS, ASA, HL

Mathematics
1 answer:
lara31 [8.8K]2 years ago
6 0

Answer:

SAS

Step-by-step explanation:

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Please please answer this correctly. Please take a pic of the same screenshot of put the right plots correctly
sergeinik [125]

Answer:

Picture

Step-by-step explanation:

I graphed them

4 0
3 years ago
Chloe reads 7 pages of a mystery book in 9 minutes. What is her average reading rate in pages per
myrzilka [38]

Answer:

1.25

Step-by-step explanation:

you have to take the 9 pages and divide by 7 and then divide that by 60 and u get 1.25

4 0
1 year ago
D^2 - 4d = 3d Solve each equation. Factor and use the zero product property to solve. Show all work algebraically.
AveGali [126]
D^2 - 4d = 3d

d^2 - 7d = 0

d(d - 7) = 0

d = 0   or   d - 7 = 0

d = 0   or   d = 7
8 0
3 years ago
Orthogonalizing vectors. Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, a
jeka57 [31]

Answer:

\\ \gamma= \frac{a\cdot b}{b\cdot b}

Step-by-step explanation:

The question to be solved is the following :

Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if b \neq 0. Recall that given two vectors a,b  a⊥ b if and only if a\cdot b =0 where \cdot is the dot product defined in \mathbb{R}^n. Suposse that b\neq 0. We want to find γ such that (a-\gamma b)\cdot b=0. Given that the dot product can be distributed and that it is linear, the following equation is obtained

(a-\gamma b)\cdot b = 0 = a\cdot b - (\gamma b)\cdot b= a\cdot b - \gamma b\cdot b

Recall that a\cdot b, b\cdot b are both real numbers, so by solving the value of γ, we get that

\gamma= \frac{a\cdot b}{b\cdot b}

By construction, this γ is unique if b\neq 0, since if there was a \gamma_2 such that (a-\gamma_2b)\cdot b = 0, then

\gamma_2 = \frac{a\cdot b}{b\cdot b}= \gamma

6 0
3 years ago
Please help due today!!!!!
fgiga [73]
The correct is the second choice
4 0
3 years ago
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