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Westkost [7]
3 years ago
10

*PLEASE HELLLPPPP ME ASAP* I HAVE BEEN STUCK ON THIS FOR DAYS

Mathematics
1 answer:
Kruka [31]3 years ago
5 0
It would be +1 -1 because it goes down 1 every time
You might be interested in
2 1/2 1 1/16 = 3 1/16 3 1/2 3 9/16 3 3/16
amm1812
2 1/2     = 2 8/16

so

2 8/16  + 1 1/16    = 3 9/16

The answer is the third option



3 0
3 years ago
Hey please help me please help please
Airida [17]

1.

a) metres to centimetres :

multiply length by 100

b) metres to millimetres:

multiply length by 1000

c) kilograms to grams:

multiply the mass value by 1000

d) litres to millilitres :

multiply volume by 1000

2.

a) 3 m = 3× 100 = 300 cm

b) 28 cm = 28 × 10 = 280 mm

c) 2.4 km = 2.4 × 1000

= 24 × 10^-1 × 10^3

= 24 × 10^2 =2400 m

d) 485 mm =485 / 10

= 485 / 10 ^1

= 485 × 10 ^-1

= 48.5 cm

e) 35 cm = 35 / 100

= 35 /10^2

= 35 × 10 ^ -2

= 0.35 m

f) 2.4 m = 2.4 / 1000

= 24 × 10 ^-1 / 10^3

= 24 × 10^-1 × 10 ^-3

= 24 × 10 ^ -4

= 0.0024 km

g) 2495 mm = 2495 /1000

= 2495 /10^ 3

= 2495 × 10 ^-3

=2.495 m

4 0
3 years ago
ASAP!: can anyone solve three times the difference of Federico age and 4 increased by 7 is greater than 37 what are possible val
Alexandra [31]

Answer:

Federcio's age possible values are those that are greater than 14 i.e 15, 16, 17 and so on

Step-by-step explanation:

Let Federico's age = x

Three times the difference of Federico’s age, and 4 means:

3(x - 4)

Increased by 7:

3(x - 4) + 7

Is greater than 37:

3(x - 4) + 7 > 37

Solving the inequality:

3x - 12 + 7 > 37

3x - 5 > 37

3x > 37 + 5

3x > 42

x >42/3

x > 14

Federcio's age possible values are those that are greater than 14

7 0
2 years ago
90°<br> 46°<br> Find the unknown. angle measure<br> please help :)
Ganezh [65]

Answer:

44 Give me brainliest please!

Step-by-step explanation:

180-90-46

90-46

44

5 0
2 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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