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2 years ago

Please help its USA Test Prep and i need an answer please and thank you

Mathematics

2 answers:

sdas [7]2 years ago

6 0

Answer:

C. 8

Step-by-step explanation:

Soloha48 [4]2 years ago

3 0

Answer:

C 8

Step-by-step explanation:

320/40 = 8

it can also work as:

320:40 divided by 40 =

8:1

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noah was assigned to make 64 cookies for the bake sale he made 125% of that number 90% of the cookies he made were sold.how many

Mashcka [7]

Answer: 8 where left after the sale

Step-by-step explanation:25% of 64 is 16, 64 plus 16 = 80, 90% of 80 is 72 so 80-72=8cookies were left.

7 0

2 years ago

Solve the systems of equation by substitution method. 2x+y=1 3x-y=4

SCORPION-xisa [38]

Answer:

x=1, y=-1. (1, -1).

Step-by-step explanation:

2x+y=1

3x-y=4

-------------

y=3x-4

2x+3x-4=1

5x-4=1

5x=1+4

5x=5

x=5/5

x=1

y=3(1)-4=3-4=-1

5 0

3 years ago

Graph

Rus_ich [418]

C is what I think/Know for sure

4 0

2 years ago

Complete the statement. Round to the nearest hundredth if necessary. 6 lb = *blank* oz

skad [1K]

Answer:

3 oz

Step-by-step explanation:

You have to divide 6 by 2 and you'll get the answer.

3 0

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

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

Thus the given are evaluated in terms of x and y

3 0

2 years ago