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d1i1m1o1n [39]
3 years ago
7

Which diagram shows possible angle measures of a triangle?

Mathematics
2 answers:
gavmur [86]3 years ago
7 0
Hello!

the angles in a triangle add to 180°
------------------------------------------------------------------------------------------------------
Triangle 1
115 + 35 + 25 = 175

Since it does not equal 180 it is not triangle 1
------------------------------------------------------------------------------------------------------
Triangle 2
115 + 32 + 33 = 180

Since it does equal 180 it is triangle 2
------------------------------------------------------------------------------------------------------
The answer is triangle 2

Hope this helps!
Deffense [45]3 years ago
4 0
The second triangle is the answer
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A furniture store pays a wholesale price for a mattress. Then, the store marks up the retail price to 150% of the wholesale pric
slega [8]
Part A:

Given that <span>the mattress is sold for 50% off of the retail price, let the retail price of the mattress be x, then

50% of x = 1200
⇒ 0.5x = 1200
⇒ x = 1200 / 0.5 = 2400

Therefore, </span><span>the retail price of the mattress, before the discount is $2,400.



Part B:

Given that </span><span>the store marks up the retail price to 150% of the wholesale price. Let the whole sale price be p, then

(100% + 150%) of p = 2400
250% of p = 2400
2.5p = 2400
p = 2400 / 2.5 = 960.

Therefore, </span><span>the wholesale price, before the markup was $960</span>
3 0
3 years ago
Simplify<br><br> (4x+2)+(x-1)=
hoa [83]

Answer:

5x+1

Step-by-step explanation:

(4x+2)+(x-1)=

Combine like terms

4x+x   +2 -1

5x+1

5 0
3 years ago
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0&lt;×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
Yep here we go again:
Shtirlitz [24]

Answer:

9 am it will be IV quadrant

11 it will be I quadrant

hope it helps u

5 0
3 years ago
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Mekhanik [1.2K]

Answer:

4

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Got it right on brainly

7 0
3 years ago
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