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Masja [62]
3 years ago
13

A ray bisects an obtuse angle into 2 congruent angles. what type of angle are those 2 congruent angles

Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
8 0
The type of angle are those 2 congruent angles is <span>acute
so your answer is </span><span>a. acute
hope that helps
</span>
VikaD [51]3 years ago
6 0
A. An obtuse angle is less than 180 but greater than 90. So for examples sake I will use 179 degrees. 179 divided by 2 (as the angles are congruent) = 89.5 which is less than a right angle. Therefore, the angles will be acute.

Hope it helps :)
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3 years ago
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6. The vertices of ALMN are L(7,4), M(7,16), and
max2010maxim [7]

Answer:

(a) LM=12 units, LN=35 units, MN=37 units

(b)8 84 units

(c) 210 square units

Step-by-step explanation:

(a)

Since points L and M have same x coordinates, it means they are in the same plane. Also, since the Y coordinates of L and N are same, they also lie in the same plane

Length LM=\sqrt {(7-7)^{2}+(16-4)^{2}}=12 units

Length LN=\sqrt {(42-7)^{2}+(4-4)^{2}}=35 units

LengthMN=\sqrt {(42-7)^{2}+(4-16)^{2}}=37 units

Alternatively, since this is a right angle triangle, length MN is found using Pythagoras theorem where

MN=\sqrt {(LN)^{2}+(LM)^{2}}=\sqrt {(12)^{2}+(35)^{2}}=37 units

Therefore, the lengths LM=12 units, LN=35 units and MN=37 units

(b)

Perimeter is the distance all round the figure

P=LM+LN+MN=12 units+35 units+37 units=84 units

(c)

Area of a triangle is given by 0.5bh where b is base and h is height, in this case, b is LN=35 units and h=LM which is 12 units

Therefore, A=0.5*12*35= 210 square units

4 0
3 years ago
Solve for x d(-3+x)= kx+9
olga nikolaevna [1]
1. Write the equation:
d(-3 + x) = kx + 9
2. "Open" the parenthesizes:
-3d + dx = kx + 9
3. Take kx to the left side, -3d to the right, everything with different sign (+ replace with - when transferring, - replace with +):
dx - kx = 9 + 3d
4. Factor x in the left side:
x(d - k) = 9 + 3d
5. Finally, divide the whole equation by (d - k):
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That's the final answer. Good luck!
4 0
4 years ago
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Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0
MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy  = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

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for u= x² → du=x*dx

I₂=  ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ =  - tan⁻¹ (x²) +C₂

then

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for y(x=1) = 0

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1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0
3 years ago
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murzikaleks [220]

Answer:

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3 0
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