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My name is Ann [436]
3 years ago
12

I need some help!lol

Mathematics
1 answer:
svetlana [45]3 years ago
7 0
Wht do you need help with exactly ?
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What is the value of x?<br><br>Enter your answer in the box. <br><br>x =​
balu736 [363]

Answer:

x=3

Step-by-step explanation:

BC=6

cos(60)=×÷6

×=3

8 0
3 years ago
Which equation has the steepest graph?anyone need some help
il63 [147K]

Answer:

Yhe equation that jas the steepest gradient is D

Step-by-step explanation:

look at the numbers infront of the 'x' without looking at the negative sign. The

larger the number the steeper the gradient. Negative gradient only means that it is a downward slope

4 0
3 years ago
21x <br> 3<br> +28x <br> 2<br> −49x−12x <br> 2<br> −16x+28
Serhud [2]

Answer:

2

1

3

+

2

8

2

−

4

9

−

1

2

2

−

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6

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Step-by-step explanation:

8 0
3 years ago
Branliets offered
Ivahew [28]
2x+20 for the first one

7 0
3 years ago
Read 2 more answers
            Find the approximate solution of this system of equations.
Montano1993 [528]
Y = |x² - 3x + 1|
y = x - 1

|x² - 3x + 1| = x - 1
|x² - 3x + 1| = ±1(x - 1)
|x² - 3x + 1| = 1(x - 1)       or      |x² - 3x + 1| = -1(x - 1)
|x² - 3x + 1| = 1(x) - 1(1)    or    |x² - 3x + 1| = -1(x) + 1(1)
|x² - 3x + 1| = x - 1        or         |x² - 3x + 1| = -x + 1
  x² - 3x + 1 = x - 1         or          x² - 3x + 1 = -x + 1
        - x        - x                                + x         + x
  x² - 4x + 1 = -1           or            x² - 2x + 1 = 1
              + 1 + 1                                       - 1 - 1
  x² - 4x + 1 = 0              or           x² - 2x + 0 = 0
  x = -(-4) ± √((-4)² - 4(1)(1))    or    x = -(-2) ± √((-2)² - 4(1)(0))
                      2(1)                                             2(1)
  x = 4 ± √(16 - 4)            or            x = 2 ± √(4 - 0)
                 2                                                 2
  x = 4 ± √(12)              or               x = 2 ± √(4)
             2                                                  2
 x = 4 ± 2√(3)               or               x = 2 ± 2
             2                                                2
 x = 2 ± √(3)                or                x = 1 ± 1
 x = 2 + √(3)  or  x = 2 - √(3)   or    x = 1 + 1    or    x = 1 - 1
                                                      x = 2       or       x = 0
y = x - 1          or           y = x - 1                            or    y = x - 1   or    y = x - 1
y = (2 + √(3)) - 1    or    y = (2 - √(3)) - 1          or         y = 2 - 1    or    y = 0 - 1
y = 2 - 1 + √(3)     or      y = 2 - 1 - √(3)          or           y = 1      or       y = -1
y = 1 + √(3)        or        y = 1 - √(3)               (x, y) = (2, 1)    or    (x, y) = (0, -1)
       (x, y) = (2 ± √(3), 1 ± √(3))

The solution (0, -1) can be made by one function (y = x - 1) while the solution (2 ± √(3), 1 ± √(3)) can be made by another function (y = |x² - 3x + 1|). So the solution (2, 1) can be made by both functions, making the two solutions equal.
4 0
3 years ago
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