All categories

3 years ago

PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPP CLICK HERE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Gwar [14]3 years ago

4 0

The answer would be a.

dexar [7]3 years ago

3 0

I think it’s A bc of how it’s set up

You might be interested in

Solve the following equation. Then place the correct number in the box provided. 6x - 8 = 16

uysha [10]

6x-8=16
6x-8+8=16+8
6x=24
6x=24
— —
6 6
X= 4

8 0

4 years ago

Read 2 more answers

Please answer!it is grade 6 math

vova2212 [387]

Answer:

AB+BE+EF+GF+HG+AH = 14 #EQU 1
BC+CD+DE+EF+GF+BG = 22 #EQU 2
AB+BC+CD+DE+EF+GF+HG+AH = 24 #EQU 3
IMPLEMENTING EQU 2 IN EQU 3
AB+22-EF+ HG+AH = 24 # EQU 5
FROM EQU 1
AB+HG+AH = 14-(BE+EF+GF) #EQU 4
IMPLEMENTING EQU 4 IN EQU 5
22+14 - (BE+EF+GF+EF) = 24
36 - 24 = BE+EF+GF+BG(EF=BG)
12 = PATH TAKEN ON THURSDAY

5 0

3 years ago

Find (-1) * (-(-2)) * (-(-(-3))) * (-(-(-(-4)))).

Brums [2.3K]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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

3 years ago

Which number represents the sum of -8 and -i? 0 87 O -87 O --8-7 0 -8 +7

mafiozo [28]

Answer:

-8+i

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers