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belka [17]
2 years ago
15

PLS HELP

Mathematics
1 answer:
Travka [436]2 years ago
6 0

Answer:

A

Step-by-step explanation:

I did it on a test too

You might be interested in
Three sodas and four hotdogs costs $15. Two sodas and a hotdog costs $5. How much does seven hotdogs cost?
KengaRu [80]

Answer:

$26.25

Step-by-step explanation:

I'm not completely sure if I did this right because I haven't used this in a while, but let's say sodas are x and hot dogs are y. The first equation would be  3x+4y=15 and in order to solve for x or y you have isolate them to seperate sides because if there is not letter on the other side, it will just end up being zero... if that makes sense. subtract 3x so the equation is now 4y=15-3x and divide everything by 4, to make the equation y=3.75-3/4x. Do the same thing but subracting the 4y and dividing by 3 to get the x value. Now we're left with y=-3/4+3.75 and x=5-4/3y (or x=5-1 1/3y). remember that x=sodas and y=hotdogs. The equation we need to create has seven hot dogs and no sodas, so we simply need to take the equation for y and substitute y for 7 because we have 7 hot dogs and y is the symbol used for hot dogs. This equation would now be 7(-3/4x+3.75) because hotdog=y=-3/4x+3.75, so hotdog=-3/4+3.75 and we have 7 hot dogs so you just multiply that by 7

7 0
2 years ago
PLEASE HELP!!!!!!!!!
Naddik [55]
The answer is A. The player runs 7 meters forward and then runs 7 meters in the opposite direction. Hope it helped!

- Shadow.
4 0
3 years ago
A line has a slope of -2/5. Generate the coordinates of two points that would lie on the same line
vovangra [49]

Answer:

Many solutions are possible. One example is (10,5) and (5,7).

Step-by-step explanation:

The slope is the distance between points represented as a ratio of vertical to horizontal distances. It is a difference or subtraction. To find two points work backwards. What two points have y values with a distance of -2?

5-7=-2

So the points are (x,5) and (x,7).

Now what x values have a distance of 5?

10-5 = 5

So the points are (10,5) and (5,7).

5 0
3 years ago
A = 32°, a = 19, b = 14<br><br> solve the triangle using the law of cos
Lesechka [4]
Here is your solution.. hope it helps

6 0
3 years ago
Read 2 more answers
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. \ }

<u>Problem (a)</u>

|4 – 3x| = |-4|

|4 – 3x| = 4

<u>Case 1</u>

\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.

<u>Case 2</u>

\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} \ }  

————————

<u>Problem (b)</u>

2|3x - 8| = 10

Divide both sides by two.

|3x - 8| = 5  

<u>Case 1</u>

\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.

<u>Case 2</u>

\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} \ }  

————————

<u>Problem (c)</u>

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

Subtracting both sides by 2x.

|3x - 8| = -2x – 4

<u>Case 1</u>

\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.

<u>Case 2</u>

\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.

————————

<u>Problem (d)</u>

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} \ }

————————

<u>Problem (e)</u>

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.

<u>Case 1</u>

  • 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.

<u>Case 2</u>

  • 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.

<u>Case 3</u>

  • 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 \ }

<u>Case 4</u>

  • 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>
  1. The inverse of a function brainly.com/question/3225044
  2. The piecewise-defined functions brainly.com/question/9590016
  3. 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
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