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NeX [460]
3 years ago
6

Please help me. If △NMK ≅ △TRP, then answer the following questions:

Mathematics
1 answer:
Natalka [10]3 years ago
3 0

Answer:

Answers are below.

Step-by-step explanation:

a. △MNK ≅ △RTP

b. TR ≅ NM

c. x = 7

hope this helps and is right!! p.s. i really need brainliest :)

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Your mother takes you to your grandparents’ house for dinner. She drives 60 minutes at a constant speed of 40 miles per hour. Sh
SIZIF [17.4K]

Answer:

A!

Step-by-step explanation:

i looked it up

5 0
3 years ago
Determine which triangle appears to be right?
anastassius [24]
A right triangle is a triangle with one 90 degree angle; the triangle that appears to have a right angle is D/ the option on the very bottom.
4 0
3 years ago
How do you solve Y=x+1 x+2y=14 in substitution?
AleksAgata [21]

Answer:

(4, 5 )

Step-by-step explanation:

Given the 2 equations

y = x + 1 → (1)

x + 2y = 14 → (2)

Substitute y = x + 1 into (2)

x + 2(x + 1) = 14 ← distribute and simplify left side

x + 2x + 2 = 14

3x + 2 = 14 ( subtract 2 from both sides )

3x = 12 ( divide both sides by 3 )

x = 4

Substitute x = 4 into (1) for corresponding value of y

y = 4 + 1 = 5

Solution is (4, 5 )

6 0
3 years ago
Use the theorem in Sec. 28 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant
alexandr1967 [171]

Answer:

The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.

Step-by-step explanation:

Given

For any given function f(z), it is analytic and not constant throughout a domain D

To Prove

The function f(z) is non-constant constant in the neighbourhood lying in D.

Proof

1-Assume that the value of f(z)  is analytic and has a constant throughout some neighbourhood in D which is ω₀

2-Now consider another function F₁(z) where

F₁(z)=f(z)-ω₀

3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.

4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.

5-Replacing value of F₁(z) in the above gives:

F₁(z)≡0 in domain D

f(z)-ω₀≡0 in domain D

f(z)≡0+ω₀ in domain D

f(z)≡ω₀ in domain D

So this indicates that the value of f(z) for all values in domain D is a constant  ω₀.

This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.

6 0
2 years ago
What is the length of the base b?
musickatia [10]

Answer:

5 units

Step-by-step explanation:

The graph helps you measure it

6 0
3 years ago
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