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kramer
4 years ago
11

How do I solve this equation??(6g^5h-4)^3

Mathematics
1 answer:
Veseljchak [2.6K]4 years ago
8 0
((6)^3 (g^5)^3 (h)^3 + (-4)^3)
(216 x g^15 x h^3 - 64
216g^15h^3 - 64
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3. Is the relation {(1,2), (3, 10), (2,8), (5,<br> 10)} a function?<br> a) yes<br> b) no<br> a
Sedaia [141]

Answer:

the answer is yes

Step-by-step explanation:

4 0
3 years ago
In which figure is DE BC
stepan [7]

Answer:

Given: DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm and DE || BC.

DB

AD

=

AC

AE

[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same.

7.2

AD

=

5.4

1.8

AD=

5.4

7.2×1.8

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10

24

⟹AD=2.4cm

4 0
3 years ago
Can someone please answer number 5?
Free_Kalibri [48]

Answer:

x=13

Step-by-step explanation:

since they are vertical angles, they equal each other so set both angles to have them equal each other to find x

8x - 12 = 6x + 14

Add 12 on both sides

8x = 6x  + 26

Subtract tax from both sides

2x = 26

Divide by 2 for both sides

x = 13

7 0
3 years ago
What is the 5:6 ratio of 330
jolli1 [7]
The ratio 5/6  is equivalent to 275/330

                 5:6   ⇔  275:330
8 0
3 years ago
Read 2 more answers
Solve triangle ABC
Goshia [24]

That figure obviously doesn't go with this problem.  It doesn't matter; this is triangle ABC labeled the usual way.

c = 10, B = 35°, C = 65°

We have two angles and a side.  The third angle is obviously

A = 180° -  35°- 65°  = 80°

The remaining sides are given by the Law of Sines,

\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}

a = \dfrac{c \sin A}{\sin C} = \dfrac{10 \sin 80}{\sin 65} = 10.866

b = \dfrac{c \sin B}{\sin C} = \dfrac{10 \sin 35}{\sin 65} = 6.328

Answer: A=80°, a=10.9, b=6.3, third choice

4 0
3 years ago
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