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juin [17]
3 years ago
6

Pleaseeeeeeeeeee help me ​

Mathematics
1 answer:
il63 [147K]3 years ago
4 0
The answer would be C
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Please can someone help with this question. I don't understand
Umnica [9.8K]

Answer:

See below,please.

Step-by-step explanation:

Volume of water = meter reading (11) - meter reading (8)

=

= 4598 - 4559 = 39 {m}^{3}

7 0
3 years ago
Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work.
dalvyx [7]

Answer:the independent variable is x, the number of hours of work.

The dependent variable is y, the total charge for x hours of work.

Step-by-step explanation:

A change in the value of the independent variable causes a corresponding change in the value of in dependent variable. Thus, the dependent variable is is output while the independent variable is the input

For each visit, he charges $25 plus $20 per hour of work. The linear expression that represents the total amount of money that Ethan earns per visit is y = 25 + 20x.

Since the total amount charged, y depends on the number of hours of work, x, it means that the dependent variable is y and the independent variable is x

4 0
3 years ago
Find the exact value of sec (–660)°
Nostrana [21]
Add full rotations of 360° degree until the angle of between 0° and 360°.
Sec (60)
The exact value is sec (60) is 2
8 0
3 years ago
Which of the following correctly justifies statement 4 of the two-column proof?
slavikrds [6]

Answer:

D.Transitive property of equality

Step-by-step explanation:

We are given that segment JK is parallel to segment LM

We have to prove \angle 3=\angle 6

We have to find which option correctly justifies the statement 4 of the two - column proof.

1.Statement : JK is parallel to segment LM

Reason: Given

2.\angle 7\cong \angle 6

Reason: Vertical angles theorem

3.\angle 3\cong \angle 7

Reason:Corresponding angles theorem

4.\angle 3\cong \angle 6

Reason: Transitive property of equality.

If a=b and b=c then a=c

Hence, option D is true.

3 0
3 years ago
Read 2 more answers
4 Tan A/1-Tan^4=Tan2A + Sin2A​
Eva8 [605]

tan(2<em>A</em>) + sin(2<em>A</em>) = sin(2<em>A</em>)/cos(2<em>A</em>) + sin(2<em>A</em>)

• rewrite tan = sin/cos

… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

… = 2/(2 cos²(<em>A</em>) - 1) (sin(<em>A</em>) cos(<em>A</em>) + sin(<em>A</em>) cos(<em>A</em>) (cos²(<em>A</em>) - sin²(<em>A</em>)))

• factor out sin(<em>A</em>) cos(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (1 + cos²(<em>A</em>) - sin²(<em>A</em>))

• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

… = 2 sin(<em>A</em>) cos(<em>A</em>)/(2 cos²(<em>A</em>) - 1) (2 cos²(<em>A</em>))

• rearrange terms in the product

… = 2 sin(<em>A</em>) cos(<em>A</em>) (2 cos²(<em>A</em>))/(2 cos²(<em>A</em>) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - 1/cos²(<em>A</em>))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

… = 4 tan(<em>A</em>) / (sec²(<em>A</em>) (2 - sec²(<em>A</em>)))

• rewrite using the Pythagorean identity, sec²(<em>A</em>) = 1 + tan²(<em>A</em>)

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (2 - (1 + tan²(<em>A</em>))))

• simplify

… = 4 tan(<em>A</em>) / ((1 + tan²(<em>A</em>)) (1 - tan²(<em>A</em>)))

• condense the denominator as the difference of squares

… = 4 tan(<em>A</em>) / (1 - tan⁴(<em>A</em>))

(Note that some of these steps are optional or can be done simultaneously)

7 0
3 years ago
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