Need help with 35 thank you so much!!
1 answer:
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Answer:
- ∠CDE ↔ 50°
- ∠FEG ↔ 75°
- ∠ACB ↔ 55°
Step-by-step explanation:
To solve angle problems like this, you make use of three relations:
- linear angles have a sum of 180°
- angles in a triangle have a sum of 180°
- vertical angles have the same measure
The attached diagram shows the measures of all of the angles of interest in the figure. The ones shown in blue are the ones that have the measures and names on the list of answer choices.
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A good place to start is with the linear angle pair at A. Since the sum of the two angles is 180°, the angle at A that is inside the triangle will be ...
180° -130° = 50°
Then the missing angle in that triangle at C will have the measure that makes the sum of triangle angles be 180°:
∠ACB = 180° -50° -75° = 55° . . . . . this is one of the angles on your list
Similarly, the angle at E inside triangle FEG will have a measure that makes those angles have a sum of 180°:
∠FEG = 180° -60° -45° = 75° . . . . . this is one of the angles on your list
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The two angles whose measures we just found are vertical angles with the base angles in triangle CDE, so that triangle's angle D will have a measure that makes the total be 180°.
∠CDE = 180° -55° -75° = 50° . . . . . this is one of the angles on your list
Answer:
Therefore, the length of the square is cm which should be cut out of each corner to get a box with the maximum volume.
Step-by-step explanation:
Given that, an open box is made from a plastic with dimension 70 cm by 96 cm by removing a square from each corner of the plastic.
Consider, x be the length of the side of the square.
After cutting the corner, the length of the plastic is = (70-2x) cm
The width of the plastic is=(96-2x) cm.
Now the length of the open box = (70-2x) cm
The width of the open box = (96-2x) cm.
Then x will be the height of the open box.
The volume of the box is
V=Length×width×height
=[(70-2x) (96-2x) x] cm³
=(6720x -192x²-140x²+4x³) cm³
=(4x³-332x²+6720x) cm³
∴V=4x³-332x²+6720x
Differentiating with respect to x
V'= 12x²-664x+6720
Again differentiating with respect to x
V''= 24x -664
To find maximum volume, we set V' =0
12x²-664x+6720=0
⇒4(3x²-166x+1680)=0
⇒3x²-166x+1680=0
⇒3x²-126x-40x+1680=0
⇒3x(x-42)-40(x-42)=0
⇒(x-42)(3x-40)=0
Now
And
At
, V has maximum value.
Therefore, the length of the square is cm which should be cut out of each corner to get a box with the maximum volume.