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Alex73 [517]
3 years ago
11

3.

Mathematics
1 answer:
steposvetlana [31]3 years ago
6 0

Answer:

Step-by-step explanation:

range?

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How do you differentiate y = 3x3 - 2x-4
riadik2000 [5.3K]

Answer:

y' = 9x^2 + 8x^-5

Step-by-step explanation:

General Principle

y' = ax^b

y' = a*b * x^(b-1)

Solution

3x^3 = 3*3x^(3-1)

3x^3 = 9x^2

-2x^-4  = -2*-4 x ^(-4 - 1)

-2x^-4 = 8 x ^-5

Answer

y' = 9x^2 + 8x^-5

7 0
3 years ago
Which of the following is a solution to the equation c+(4-3c)-2=0
Eduardwww [97]

Answer:

c=1

Step-by-step explanation:

3 0
3 years ago
PLEASE HELPPPPP
Flauer [41]

Answer:

Cost of 1 rose bush = x = $5

Cost of 1 geraniums = y = $2

Step-by-step explanation:

DeShawn and Mike each improved their yards by planting rose bushes and geraniums. They bought their supplies from the same store.

Let us represent:

Cost of 1 rose bush = x

Cost of 1 geraniums = y

DeShawn spent $13 on 1 rose bush and 4 geraniums.

x + 4y = 13..... Equation 1

x = 13 - 4y

Mike spent $56 on 10 rose bushes and 3 geraniums.

10x + 3y = 56....... Equation 2

We substitute 13 - 4y for x in Equation 2

10(13 - 4y) + 3y = 56

130 - 40y + 3y = 56

Collect like terms

- 40y + 3y = 56 - 130

-37y = -74

y = -74/37

y = $2

Solving for x

x = 13 - 4y

x = _13 - 4 × $2

x = $13 - $8

x = $5

Therefore,

Cost of 1 rose bush = x = $5

Cost of 1 geraniums = y = $2

7 0
3 years ago
Help PLEASE I need help on this question ASAP!!!!
loris [4]

Answer:

$8

Step-by-step explanation:

14 - 6 = 8

3 0
3 years ago
Read 2 more answers
Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!
blagie [28]

<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

2) \angle FBE \cong \angle GBE (an angle bisector splits an angle into two congruent parts)

3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

5) \overline{BE} \cong \overline{BE} (reflexive property)

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

5) \overline{OX} \cong  \overline{OY} (CPCTC)

<u>Question 6</u>

1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

2) \overline{AC} \cong \overline{AC} (reflexive property)

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

5) \triangle BAC \cong \triangle DCA (HA)

6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

5) \overline{AD} \cong \overline{AD} (reflexive property)

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

7) \angle BDA \cong \angle CDA (CPCTC)

8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

8 0
1 year ago
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