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3 years ago

Solve for r. r + 62 ÷2 = 1

2 answers:

7 0

I’m pretty sure it’s -30. Here’s what I did:
1. Divided 62 and 2 which gave me 31
2. Then subtract 31 and 1 which will give you -30.
I’m not the best at explaining things but I hope this helps

DENIUS [597]3 years ago

4 0

Answer:

-30 because -30 + 31 = 1.

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4 years ago

Divide £700 in the ratio 6:4 (2 marks)

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One num = 6x = 6 x 70 = 420

Second= 4x = 4 x 70 = 280

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3 years ago

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Zolol [24]

Answer:

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3 years ago

Read 2 more answers

Condense the following logs into a single log:

mamaluj [8]

QUESTION 1

The given logarithm is

$8\log_g(x)+5\log_g(y)$

We apply the power rule of logarithms; $n\log_a(m)=\log_(m^n)$

$=\log_g(x^8)+\log_g(y^5)$

We now apply the product rule of logarithm;

$\log_a(m)+\log_a(n)=\log_a(mn)$

$=\log_g(x^8y^5)$

QUESTION 2

The given logarithm is

$8\log_5(x)+\frac{3}{4}\log_5(y)-5\log_5(z)$

We apply the power rule of logarithm to get;

$=\log_5(x^8)+\log_5(y^{\frac{3}{4}})-\log_5(z^5)$

We apply the product to obtain;

$=\log_5(x^8\times y^{\frac{3}{4}})-\log_5(z^5)$

We apply the quotient rule; $\log_a(m)-\log_a(n)=\log_a(\frac{m}{n} )$

$=\log_5(\frac{x^8\times y^{\frac{3}{4}}}{z^5})$

$=\log_5(\frac{x^8 \sqrt[4]{y^3} }{z^5})$

7 0

3 years ago

Solve for r. r + 62 ÷2 = 1​

Solve for r. r + 62 ÷2 = 1