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Phantasy [73]
3 years ago
10

Help me please I give thanks!

Mathematics
2 answers:
timofeeve [1]3 years ago
6 0
If we assume that the lines BC and the ilne that pint A is on are paralell  since they didn't specify

remember at interior angles are congruent
also oppiotie angles are congruent
therefor angle x is congruent to 3x-70
so solve
x=3x-70
add 70 to both sides
70+x=3x
subtract x from both sides
70=2x
divide by 2
35=x
angle abc needs to be found
since that euals angle x
the asnwer is 35 degrees
professor190 [17]3 years ago
4 0
Because the lines are parallel the value of the angle is the same

x=3x-70
-2x =-70
2x=70
x=35
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tamaranim1 [39]

Answer:

B

Step-by-step explanation:

The pattern is x/-3

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Read 2 more answers
Points of discontinuity.<br> y = cot x, [0, 2 \[\pi \] ]
Triss [41]
<span>cotx = (cosx)/(sinx)
 Where is this function not defined?
i.e. where is sin(x)=0?
Because at this point, cotx will be undefined because you cannot divide by zero. So find all points from 0 to 2 pi where sinx is 0.</span>
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3 years ago
Solve the equation.<br>-Эх + 1 = -x+ 17​
makkiz [27]

Answer:

x = -8. solve by balancing the equation.

Step-by-step explanation:

-2x = 16

x = -8

5 0
3 years ago
HELP NOW PLZ 20 POINTS FOR BRAINLIEST Step 1: Write a quadratic equation in standard form choosing your own coefficients and con
Gwar [14]

Answer:

Here's what I get  

Step-by-step explanation:

y = ax² + bx + c

1. Equation 1

I will use the equation

y = x² + 2x - 3

a = 1; b= 2; c = -3

a = 1, so the parabola opens up.

c < 0, so the y-intercept is negative.

Strategy:

I will factor and use the zero-product property because

3(-1) = -3 and 3 + (-1) = 2

(a) Find two numbers that multiply to give ac (13), and add to give b (2).

Factors of 2: +1,±2

Factors of 3: ±1, ±3

After some trial and error, you find the numbers are 3 and -1.

3 - 1 = 2 and 3(-1) = -3

(b) Rewrite the middle with those numbers

x² + 3x + (-1)x - 3

(c) Factor the first two and last two terms separately

x(x +3) + (-1)(x + 3)

(d) Take out the common factor

(x + 3)(x - 1)

(e) Apply the zero-product property

\begin{array}{rcl}x + 3 = 0 & \qquad & x - 1 = 0\\x = \mathbf{-3} & \qquad & x = \mathbf{1}\\\end{array}

2. Equation 2

y = -x² + 2x - 1

a = -1; b = 2; c = -1

a = -1, so the parabola opens down.

c < 0, so the y-intercept is negative.

Strategy:

I will solve this one by completing the square because it is less obvious how to factor a quadratic with a negative leading coefficient.

-x² + 2x - 1 = 0

(a) Divide each side by -1

x² - 2x + 1 = 0

(b) Move the constant term to the right-hand side

x² - 2x = -1

(c) Take half the coefficient of x, square it, and add it to each side

-2/2 = 1

   1² = 1

x² - 2x +1 = 0

(d) Factor each side as the square of a binomial

(x - 1)² = 0

(e) Take the square root of each side

x - 1 = 0

(f) Solve for x

x = 1

3. Equation 3

y = -½x² + 1.9x + 3

a = -½; b = 1.9; c = +3

a = -½, so the parabola opens down and is compressed vertically.

c = 2, so the y-intercept is at y = 3

Strategy:

We could I will solve this one graphically because the fractional coefficients look problematic.

(a) Make a table containing a few points

\begin{array}{rr}\mathbf{x} & \mathbf{y} \\-2 & -2.8 \\0 & 3.0 \\2 & 4.8 \\4 & 2.6 \\6&-3.6\end{array}

(b) Plot the points

(c) Join the points with a smooth curve

(d) Extend the curve to the edges of  the graph

The graph below shows your calculated points and the x-intercepts at  

x = -1.2 and x = 5

4. Equation 4

2x² - 8 = 0

a = 2; b = 0; c = -8

a = 2, so the parabola opens upward and is vertically stretched.

c = -8, so the y-intercept is at y = -8.

Strategy:

b = 0, so I will use the method of taking the square root of each side.

(a) Remove the common factor

2x² - 8 = 0

  x² - 4 = 0

<h3>(b) Move the constant term to the other side </h3>

x² = 4

(c) Take the square root of each side

x = ±2

5. Equation 5

y = x² + 3x + 1

This equation has no rational factors, so the formula will be useful.

a = 1; b = 1, c = 1

\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\\\ & = & \dfrac{-3\pm\sqrt{3^2 - 4\times 1\times 1}}{2\times 1} \\\\ & = & \dfrac{-3\pm\sqrt{9 - 4}}{-2} \\\\ & = & \dfrac{-3\pm\sqrt{5}}{-2} \\\\x=\dfrac{-3 + \sqrt{5}}{-2} &   &x=\dfrac{-3 - \sqrt{5}}{-2}\\\\x=\mathbf{\dfrac{3}{2} - \dfrac{\sqrt{5}}{2}}& &   x = \mathbf{\dfrac{3}{2} + \dfrac{\sqrt{5}}{2}}\\\\\end {array}

5 0
3 years ago
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