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3 years ago

Solve for x x-4 — = 5 y+2

Mathematics

1 answer:

yulyashka [42]3 years ago

5 0

I think the question is wrong!

It must be

x-4

___= 5

x+2

Answer: x - 4= 5x + 10

or,-10 -4 = 5x - x

or, -14=4x

or,-14/4= x

Therefore, x = -14/4

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Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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3) 0.07 x 16<br> Help now

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Answer:

1.12

Step-by-step explanation:

all you have to do is multiply them, hope i helped

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3 years ago

Read 2 more answers

Betty has a bag with 20 marbles in it. Some of the marbles are blue, some are green, and some are yellow.

Nezavi [6.7K]

392 / 500 double and take one decimal place to see percentage

52/500

56/500 e.g. 56*2 = 112 now take 1 decimal so it is 11.2%

now divide each by 20

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3 years ago

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Mike has $6000 to invest. He invested part of his total into an account earning 6% interest and the rest in an account earning 1

lesya692 [45]

Answer:$4000 was invested into the account earning 6% interest.

$2000 was invested into the account earning 11% interest.

Step-by-step explanation:

Let x represent the amount invested into the account earning 6% interest.

Let y represent the amount invested into the account earning 11% interest.

Mike has $6000 to invest. He invested part of his total into an account earning 6% interest and the rest in an account earning 11% interest. This means that

x + y = 6000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Considering the account earning 6% interest, the interest would be

I = (x × 6 × 1)/100 = 0.06x

Considering the account earning 11% interest, the interest would be

I = (x × 11 × 1)/100 = 0.11y

If at the end of the year, he earned $460.00 in interest, it means that

0.06x + 0.11y = 460 - - - - - - - - - - -1

Substituting x = 6000 - y into equation 1, it becomes

0.06(6000 - y) + 0.11y = 460

360 - 0.06y + 0.11y = 460

- 0.06y + 0.11y = 460 - 360

0.05y = 100

y = 100/0.05 = $2000

x = 6000 - y = 6000 - 2000

x = $4000

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Create<br> height<br> question about<br> Grade 7

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Answer:

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Step-by-step explanation:

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