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3 years ago

What is the answer to -5x+5y=-10 and 3x-3y=5 using substitution

Mathematics

1 answer:

IceJOKER [234]3 years ago

7 0

Answer:

for the 1st one the answer is x= 3 y=1 and the second one is x= 5 and y = 3 and 1/3

Step-by-step explanation: so basically for the first one -5 times x which is 3 is -15 right? and then 5 times y which is 1 is 5 right so -15 plus 5 = -10

second one is x = 5 and y = 3 and 1/3 because 5*x which is 5 is = to 15 right? and then 3 times y which is equal to 3 and 1/3 is equal to 10 which when you do 15-10 in pretty sure it equals 5 correct me if im wrong

Hope this helps xD ;)

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3 years ago

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Step-by-step explanation:

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3 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 $49^{(2t - 0.5)}$

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3 years ago

What is the answer to 2x+4y=16:4y=12

allsm [11]

Answer:

Here,

$2x + 4y = 16....eqn \: i \\ 4y = 12....eqn \: ii$

substituting the value of 4y in eqn i fron eqn ii

so eqn i becomes

$2x + 12 = 16 \\ 2x = 4 \\ x = 2$

$y = 12 \div 4 = 3$

Thus,

$x = 2 \: and \: y = 3$

...hope this helps:)

8 0

1 year ago