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3 years ago

Adam will plant up to 41 acres on his farm with wheat and corn. Fewer than 28 acres will be planted with corn.

Mathematics

2 answers:

Eddi Din [679]3 years ago

8 0

Both A and D are correct. Yup.

Vlad1618 [11]3 years ago

6 0

Answer:

w+c ≤ 41

c < 28

Step-by-step explanation:

Given :Adam will plant up to 41 acres on his farm with wheat and corn. Fewer than 28 acres will be planted with corn.

To Find: Identify two inequalities that represent this situation.

Solution:

We are given that

w represent the number of acres of wheat

c represent the number of acres of corn.

since we are given that Adam will plant up to 41 acres on his farm with wheat and corn.

⇒w+c ≤ 41

We are also given that Fewer than 28 acres will be planted with corn.

⇒c < 28

Hence Option A and D is correct.

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9 + x - 3 = 4x - 3 Find x, please explain asap

Vesnalui [34]

Answer:

x = 3

Explanation:

To simplify this equation, don't worry about the -3, since there is one on both sides. Therefore, they cancel out.

So, 9 + x = 4x

When you plug 3 in for x, it becomes:

9 + 3 = 4 x 3

12 = 12

This equation is true, therefore, x is your answer!

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2 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$