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3 years ago

PLEASE HELP ME ?!!!!!

Mathematics

1 answer:

cluponka [151]3 years ago

8 0

Answer:

b & c for sure

Step-by-step explanation:

............

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Jason ran 5/7 of the distance around the school track. Sara ran 4/5 of Jason's distance. What fraction of the total distance aro

masha68 [24]

Jason: 5/7
Sara: 4/5

Think of it this way: if the total track is a mile, and Jason runs 5/7 of it, he has run 5/7 of a mile. Then Sara runs 4/5 OF 5/7, with "of" meaning "times," so she runs 4/5 x 5/7, which gives you 20/35, which simplifies to 4/7. You can also think of Sara's distance as 80 percent of Jason's. If she runs "80 percent of the sevenths that Jason ran," that means she ran 4 out of his 5 sevenths.

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3 years ago

A system of two equations contains one quadratic equation and one linear equation. the quadratic system of the equation is y=x^2

ruslelena [56]

Suppose the line is y=mx+b
when x=3, y=15, so 15=3m+b
when x=-1, y=-13, so -13=-m+b
solve the system of these two equation, you get m=7, b=-6
so the linear equation is y=7x-6

7 0

3 years ago

Hitunglah nilai x ( jika ada ) yang memenuhi persamaan nilai mutlak berikut . Jika tidak ada nilai x yang memenuhi , berikan ala

Julli [10]

(a). The solutions are 0 and ⁸/₃.

(b). The solutions are 1 and ¹³/₃.

(c). The equation has no solution.

(d). The only solution is ²¹/₂₀.

(e). The equation has no solution.

<h3>Further explanation</h3>

These are the problems with the absolute value of a function.

For all real numbers x,

$\boxed{ \ |f(x)|=\left \{ {{f(x), for \ f(x) \geq 0} \atop {-f(x), for \ f(x) < 0}} \right. \ }$

Problem (a)

|4 – 3x| = |-4|

|4 – 3x| = 4

Case 1

$\boxed{ \ 4 - 3x \geq 0 \ } \rightarrow \boxed{ \ 4\geq 3x \ } \rightarrow \boxed{ \ x\leq \frac{4}{3} \ }$

For 4 – 3x = 4

Subtract both sides by four.

-3x = 0

Divide both sides by -3.

x = 0

Since $\boxed{ \ 0\leq \frac{4}{3} \ }$ , x = 0 is a solution.

Case 2

$\boxed{ \ 4 - 3x < 0 \ } \rightarrow \boxed{ \ 4 < 3x \ } \rightarrow \boxed{ \ x > \frac{4}{3} \ }$

For -(4 – 3x) = 4

-4 + 3x = 4

Add both sides by four.

3x = 8

Divide both sides by three.

$x = \frac{8}{3}$

Since $\boxed{ \ \frac{8}{3} > \frac{4}{3} \ }$ , $\boxed{ \ x = \frac{8}{3} \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 0 \ and \ \frac{8}{3} \ }$

————————

Problem (b)

2|3x - 8| = 10

Divide both sides by two.

|3x - 8| = 5

Case 1

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x - 8 = 5

Add both sides by eight.

3x = 13

Divide both sides by three.

$x = \frac{13}{3}$

Since $\boxed{ \ \frac{13}{3} \geq \frac{4}{3} \ }$ , $\boxed{ \ x = \frac{13}{3} \ }$ is a solution.

Case 2

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x – 8) = 5

-3x + 8 = 5

Subtract both sides by eight.

-3x = -3

Divide both sides by -3.

x = 1

Since $\boxed{ \ 1 < \frac{8}{3} \ }$ , $\boxed{ \ x = 1 \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 1 \ and \ \frac{13}{3} \ }$

————————

Problem (c)

2x + |3x - 8| = -4

Subtracting both sides by 2x.

|3x - 8| = -2x – 4

Case 1

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x – 8 = -2x – 4

3x + 2x = 8 – 4

5x = 4

$x = \frac{4}{5}$

Since $\boxed{ \ \frac{4}{5} \ngeq \frac{8}{3} \ }$ , $\boxed{ \ x = \frac{4}{5} \ }$ is not a solution.

Case 2

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x - 8) = -2x – 4

-3x + 8 = -2x – 4

2x – 3x = -8 – 4

-x = -12

x = 12

Since $\boxed{ \ 12 \nless \frac{8}{3} \ }$ , $\boxed{ \ x = 12 \ }$ is not a solution.

Hence, the equation has no solution.

————————

Problem (d)

5|2x - 3| = 2|3 - 5x|

Let’s take the square of both sides. Then,

[5(2x - 3)]² = [2(3 - 5x)]²

(10x – 15)² = (6 – 10x)²

(10x - 15)² - (6 - 10x)² = 0

According to this formula $\boxed{ \ a^2 - b^2 = (a + b)(a - b) \ }$

$[(10x - 15) + (6 - 10x)][(10x - 15) - (6 - 10x)]] = 0$

(-9)(20x - 21) = 0

Dividing both sides by -9.

20x - 21 = 0

20x = 21

$x = \frac{21}{20}$

The only solution is $\boxed{ \ \frac{21}{20} \ }$

————————

Problem (e)

2x + |8 - 3x| = |x - 4|

We need to separate into four cases since we don’t know whether 8 – 3x and x – 4 are positive or negative. We cannot square both sides because there is a function of 2x.

Case 1

8 – 3x is positive (or 8 - 3x > 0)
x – 4 is positive (or x - 4 > 0)

2x + 8 – 3x = x – 4

8 – x = x – 4

-2x = -12

x = 6

Substitute x = 6 into 8 – 3x ⇒ 8 – 3(6) < 0, it doesn’t work, even though when we substitute x = 6 into x - 4 it does work.

Case 2

8 – 3x is positive (or 8 - 3x > 0)
x – 4 is negative (or x - 4 < 0)

2x + 8 – 3x = -(x – 4)

8 – x = -x + 4

x – x = = 4 - 8

It cannot be determined.

Case 3

8 – 3x is negative (or 8 - 3x < 0)
x – 4 is positive. (or x - 4 > 0)

2x + (-(8 – 3x)) = x – 4

2x – 8 + 3x = x - 4

5x – x = 8 – 4

4x = 4

x = 1

Substitute x = 1 into 8 - 3x, $\boxed{ \ 8 - 3(1) \nless 0 \ }$ , it doesn’t work. Likewise, when we substitute x = 1 into x – 4, $\boxed{ \ 1 - 4 \not> 0 \ }$

Case 4

8 – 3x is negative (or 8 - 3x < 0)
x – 4 is negative (or x - 4 < 0)

2x + (-(8 – 3x)) = -(x – 4)

2x – 8 + 3x = -x + 4

5x + x = 8 – 4

6x = 4

$\boxed{ \ x=\frac{4}{6} \rightarrow x = \frac{2}{3} \ }$

Substitute $x = \frac{2}{3} \ into \ 8-3x$ , $\boxed{ \ 8 - 3 \bigg(\frac{2}{3}\bigg) \not< 0 \ }$ , it doesn’t work. Even though when we substitute $x = \frac{2}{3} \ into \ x-4$ , $\boxed{ \ \bigg(\frac{2}{3}\bigg) - 4 < 0 \ }$ it does work.

Hence, the equation has no solution.

<h3>Learn more</h3>

The inverse of a function brainly.com/question/3225044
The piecewise-defined functions brainly.com/question/9590016
The composite function brainly.com/question/1691598

Keywords: hitunglah nilai x, the equation, absolute value of the function, has no solution, case, the only solution

5 0

3 years ago

Read 2 more answers

25) PLEASE HELP WITH QUESTION, WILL MARK BRAINLIEST + POINTS.

iogann1982 [59]

You take the three zeros and put them back with the x s

(x-6)(x+5)(x-2) they are reversed when you put them back in

multiply the first two and get $x^2+5x -6x-30(x-2)$
simplify to $x^2-x-30(x-2)$

Multiply it out again and get $x^3-x^2-30x-2x^2+2x+60$
finally simplify and get $x^3-3x^2-28x+60$

So the answer is the first one

4 0

3 years ago

How much will it cost in Canadian dollars to purchase €2000 at a bank that charges a 2.2% commission on the transaction? Use an

Ivanshal [37]

Answer:

$2,874.25 Canadian dollar

Step-by-step explanation:

The first thing to do here is to calculate the amount of the commission.

That would be 2.2% of €2,000

= 2.2/100 * €2,000 = €44

Now the total cost to pay = selling price + amount in commissions = €2,000 + €44 = €2,044

But we need our answer in the Canadian dollars

Mathematically;

$1 Canadian = €0.711142

$x Canadian = €2,044

x = (2,044 * 1)/0.711142

x = 2,874.25014975912

This is approximately $2,874.25 Canadian dollar

4 0

3 years ago