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grin007 [14]
3 years ago
7

If DEF ~= PQR, which congruences are true by CPCTC? Select all that apply.

Mathematics
1 answer:
tatyana61 [14]3 years ago
6 0

Answer:

A, C, E, F

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

8 0
3 years ago
Solve the system using combination method to get a solution (x,y)<br> 4x+7y=10<br> -4x-6y=28
PIT_PIT [208]
\left \{ {{4x+7y=10} \atop {-4x-6y=28}} \right. \\\\add\ those\ two\ equations\\\\&#10;y=38\\\\&#10;4x+7y=10\ \ \ | subtract\ 7y\\\\&#10;4x=10-7x\ \ \ \ | divide\ by\ 4\\\\&#10;x=\frac{10-7x}{4}=\frac{10-7*38}{4}=\frac{10-266}{4}=\frac{-256}{4}=-64\\\\&#10;solution:\\ \left \{ {{y=38} \atop {x=-64}} \right.
3 0
3 years ago
Read 2 more answers
Write the equation of a line in slope intercept form that has a slope of -½ and contains (4, –5)
evablogger [386]

Answer:

Step-by-step explanation:

y + 5 = -1/2(x - 4)

y + 5 = -1/2x + 2

y = -1/2x - 3

6 0
3 years ago
The price of a share of stock started the day at $37. During the day it went down $3, up $1, down $7 and up $4. What was the pri
Oksana_A [137]
It started at $37.

It went down $3.
37 - 3 = 34

It went up 1.
34 + 1 = 35

It went down 7.
35 - 7 = 28

It went up 4.
28 + 4 = 32

The price at the end of the day was $32.
4 0
3 years ago
a 10 gram sample of a substance that's used to detect explosives has k-value of 0.1353, find the substance half life in days, Ro
IrinaK [193]
We find the value of N₀ since we are provided with initial conditions.
The condition is that, at time t = 0, the amount of substance contains originally 10 grams.
We substitute:
10 = N₀ (e^(-0.1356)*0)
10 = N₀ (e^0)
N₀ = 10

When the substance is in half-life (meaning, the half of the original amount), it contains 5 grams. We solve t in this case.
5 = 10 e^(-0.1356*t)
0.5 = e^(-0.1356*t)
Multiply natural logarithms on both sides to bring down t.
ln(0.5) = -0.1356*t
Hence,
t = -(ln(0.5))/0.1356
t ≈ 5.11 days (ANSWER)
4 0
3 years ago
Read 2 more answers
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