All categories

3 years ago

One number is five more than two times another. If there sum is decreased by one, the result is sixteen. Find the numbers.

1 answer:

6 0

First no. = x
Second no. = y

Equation Formation.
x = 2 y + 5
x - 2 y = 5.........equation (i)

x + y - 1 = 16
x + y = 17............. equation(ii)

Solving Equation
x + y = 17
x - 2 y = 5
⇅
3 y = 12
y = 4

x = 2 y +5
x = 2(4) + 5
x = 13

Answer = 4 , 13

You might be interested in

I’m struggling with this question, please help me

vichka [17]

Answer:

Step-by-step explanation:

8 0

2 years ago

If log7=a and log…<br> help me solve ths math problem on picture

tatiyna

$\qquad\qquad\huge\underline{\boxed{\sf Answer☂}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: log_{4}(39.2)$

$\qquad \sf \dashrightarrow \: log_{4}( {2}^{3} \times {7}^{2} \times 10 {}^{ - 1} )$

$\qquad \sf \dashrightarrow \: log_{4}( {2}^{3} ) + log_{4}( {7}^{2} ) + log_{4}(10 {}^{ - 1} )$

$\qquad \sf \dashrightarrow \: log_{ {2}^{2} }( {2}^{3} ) + log_{ {2}^{2} }( {7}^{2} ) + log_{ {2}^{2} }(10 {}^{ - 1} )$

$\qquad \sf \dashrightarrow \: \frac{3}{2} log_{2}(2) + \frac{2}{2} log_{2}(7) - \frac{1}{2} log_{2}(10)$

$\qquad \sf \dashrightarrow \: \frac{3}{2} + a - \frac{b}{2}$

That's the required result ~

also you can take it's LCM and write it as :

$\qquad \sf \dashrightarrow \: \dfrac{2a - b + 3}{2}$

3 0

1 year ago

Y=|4x|-1 what is the absolute value

maks197457 [2]

Answer:

-1

Step-by-step explanation:

5 0

3 years ago

Given that 3x – 4y= 22<br> -<br> Find & when y = -7

Liono4ka [1.6K]

<h2>Answer:</h2>

x=2

<h2><em><u>Step-by-step explanation:</u></em></h2>

6 0

2 years ago

Determine whether each expression is equivalent to 49^2t – 0.5.

vampirchik [111]

Answer:

None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

$7^{(4t - 1)}$

This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

This is not equivalent to $49^{(2t - 0.5)}$

B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{7^{4t}}{7^{0.5}}$ = $7^{4t - 0.5}$

This is also not equivalent to $49^{(2t - 0.5)}$

C. $7^{2t}\ *\ 49^{0.5}$

$7^{2t}\ *\ (7^2)^{0.5}$

$7^{2t}\ *\ 7^{2*0.5}$

$7^{2t}\ *\ 7^{1}$

Using law of indices which states;

$a^m*a^n = a^{m+n}$

$7^{2t+ 1}$

This is also not equivalent to $49^{(2t - 0.5)}$

6 0

3 years ago