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3 years ago

Question 5) Which of the following equations is equivalent to 5 - 2x + 7y = 4 ?

Mathematics

2 answers:

Ann [662]3 years ago

4 0

Answer:

Step-by-step explanation:

Given

5 - 2x + 7y = 4 ( subtract 5 from both sides )

- 2x + 7y = - 1 ( add 2x to both sides )

7y = 2x - 1 ( divide all terms by 7 )

y = $\frac{2}{7}$ x - $\frac{1}{7}$ → B

Lilit [14]3 years ago

4 0

Answer:

B. $y=\frac{2}{7}x-\frac{1}{7}$

Step-by-step explanation:

We have been given an equation $5-2x+7y=4$ . We are asked to choose the equation that is equivalent to our given equation.

We will convert our given equation in point-slope form of equation.

$5-2x+2x+7y=4+2x$

$5+7y=4+2x$

$5-5+7y=4-5+2x$

$7y=-1+2x$

$7y=2x-1$

Divide both sides by 7:

$\frac{7y}{7}=\frac{2x}{7}-\frac{1}{7}$

$y=\frac{2}{7}x-\frac{1}{7}$

Therefore, equation $y=\frac{2}{7}x-\frac{1}{7}$ is equivalent to our given equation and option B is the correct choice.

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Use grouping symbols and at least one exponent to write a numerical expression that has a value of 80.

Kaylis [27]

To group an expression, means that the factors of the expression are to be separated in (), [] or {}.

An expression with a value of 80 is: $2^2 \times ( [2 + 2] \times [3 + 2] )$

The number is given as:

$80$

Factorize

$80 = 2 \times 40$

Factorize 40

$80 = 2 \times 2 \times 20$

Express $2 \times 2$ as an exponent

$80 = 2^2 \times 20$

Express 20 as 4 x 5

$80 = 2^2 \times ( 4 \times 5)$

Express 5 as 3 + 2

$80 = 2^2 \times ( 4 \times [3 + 2] )$

Express 4 as 2 + 2

$80 = 2^2 \times ( [2 + 2] \times [3 + 2] )$

Hence, the above expression has a numerical value of 80

Read more about numerical expressions at:

brainly.com/question/24787226

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2 years ago

Which fraction is equivalent to negative 4 over -9

boyakko [2]

I hope this helps you

8 0

3 years ago

Two 120 resistors are connected in parallel across voltage supply of 10 vdc what’s the current in this circuit

Daniel [21]

Answer: 0.167

Step-by-step explanation:

This is an applied mathematics question

Resistors in parallel ⁾/R = ¹/₁₂₀ + ¹/₁₂₀ since the resistor is two 120

Therefore , resolve into fraction now yield

¹/R = 1 + 1

-------

120

= 2/120

= ¹/₆₀

Therefore, ¹/R = ¹/₆₀

Cross multiply to get R the resistance

R = 60

From formula, V = IR where I is the current

There fore making I the subject of the formula

I = V/R

= 10/60

= 0.166

= 0.167 A, A = ampere which is the unit of current.

4 0

3 years ago

Find the mean, mode and median of 3, 4, 5, 4, 6, 4, 7.

Luda [366]

Mode = is the number with the highest frequency so the answer is 4.
Mean = the average so the answer is 3+4+5+4+6+4+7=33 so 33/7=4.7 (rounded to the nearest one decimal).
Median = the middle number, so the answer is 4.

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2 years ago

We select n + 1 diﬀerent integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte

just olya [345]

Answer:

See answer below

Step-by-step explanation:

From the set

{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.

Then

it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers

if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3 → the largest common divisor of that number with 3 is 1

If the set includes any other prime number → the largest common divisor of that with any other is 1

For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :

N = N₂*n₂ , since n₂ ≥ 2 → N₂ < N

then the even N₂ also should be contained in the set

therefore also for N₂

N = N₃*n₃ → N₃ < N₂

therefore if we continue , we would obtain a number even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1

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4 years ago