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3 years ago

: £50 is worth €56 How many euros is £1 worth?

Mathematics

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

Let's make this a ratio:

50:56

Divide by 50 on both sides (since they made it a unit rate).

1:1.12

1.12 is your answer.

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Someone please help asap

andrew-mc [135]

Answer:

the second one

Step-by-step explanation:

because g(x) and f(x) is bout the same

6 0

3 years ago

B2 is A= 1/2h(bl+b2), if A=16, h=4, and b1=3<br> b2=____

Yuliya22 [10]

Answer:

b₂ = 5

Step-by-step explanation:

Substitute the given values into the formula

16 = $\frac{1}{2}$ × 4 × (3 + b₂)

16 = 2(3 + b₂) ( divide both sides by 2 )

8 = 3 + b₂ ( subtract 3 from both sides )

5 = b₂

7 0

3 years ago

3y + 1.8 ≥ 10.8 I need mad help on dis

soldier1979 [14.2K]

Answer:

In this question the <em>y</em> is 3.

Step-by-step explanation:

3 x 3 is 9.

9 + 1.8 = 10.8

3 0

3 years ago

6÷0.9 help plzzzzzzzzzzzzzzzzzzz

Rus_ich [418]

6 / .9 *10/10

60/9

9 goes into 60 6 times with 6 left over

6 6/9

6 2/3

8 0

4 years ago

Read 2 more answers

Prove-: sin5A = 5cos^4 A sinA - 10cos^2 A sin^3 A + sin^5 A

jeyben [28]

$A=x\\\\sin5x=5cos^4xsinx-10cos^2xsin^3x+sin^5x\\\\L=sin(3x+2x)=sin3xcos2x+sin2xcos3x=(*)\\-------------------------------\\sin3x=sin(2x+x)=sin2xcosx+sinxcos2x\\=2sinxcosxcosx+sinx(cos^2x-sin^2x)\\=2sinxcos^2x+sinxcos^2x-sin^3x=3sinxcos^2x-sin^3x\\-------------------------------$

$sin3xcos2x=(3sinxcos^2x-sin^3x)(cos^2x-sin^2x)\\=3sinxcos^4x-3sin^3xcos^2x-sin^3xcos^2x+sin^5x\\=3sinxcos^4x-4sin^3xcos^2x+sin^5x\\-------------------------------$

$cos3x=cos(2x+x)=cos2xcosx-sin2xsinx\\=cosx(cos^2x-sin^2x)-2sinxcosxsinx\\=cos^3x-sin^2xcosx-2sin^2xcosx=cos^3x-3sin^2xcosx\\-------------------------------$

$sin2xcos3x=2sinxcosx(cos^3x-3sin^2xcosx)\\=2sinxcos^4x-6sin^3xcos^2x\\-------------------------------$

$
(*)=3sinxcos^4x-4sin^3xcos^2x+sin^5x+2sinxcos^4x-6sin^3xcos^2x\\\\=5sinxcos^4x-10sin^3xcos^2x+sin^5x=R$

$=======================================\\\\sin(\alpha+\beta)=sin\alpha cos\beta+sin\beta cos\alpha\\\\cos(\alpha+\beta)=cos\alpha cos\beta-sin\alpha sin\beta\\\\sin2\alpha=2sin\alpha cos\alpha\\\\cos\alpha=cos^2\alpha-sin^2\alpha$

5 0

3 years ago