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3 years ago

Perpendicular lines AB and CD intersect at point E. If m2AED = x+ 20, what is the value of

Mathematics

1 answer:

Leona [35]3 years ago

5 0

Answer:

x = 70

Step-by-step explanation:

Since the lines are perpendicular then ∠ AED = 90°, thus

x + 20 = 90 ( subtract 20 from both sides )

x = 70

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A) by completing the square find in the form of the constant p the roots of the equation x^2+px+4=0

iris [78.8K]

Answer:

a)

$\displaystyle x=-\frac{1}{2}p\pm\sqrt{\frac{1}{4}p^2-4}$

b) p is in the interval (-4,4)

Step-by-step explanation:

Quadratic Equation

It's given the following quadratic equation:

$x^2+px+4=0$

It's required to complete squares and find the roots. This can be done by recalling the polynomial identity:

$a^2+2ab+b^2=(a+b)^2$

We already have the first term squared, and we need to find the second term. Rewriting the equation:

$\displaystyle x^2+2(\frac{1}{2}px)+4=0$

The second term of the binomial is 1/2p, thus completing the squares with $b^2$ :

$\displaystyle x^2+2(\frac{1}{2}px)+\left(\frac{1}{2}p\right)^2+4-\left(\frac{1}{2}p\right)^2=0$

Factoring:

$\displaystyle \left(x+\frac{1}{2}p\right)^2+4-\frac{1}{4}p^2=0$

Moving the independent term to the right side:

$\displaystyle \left(x+\frac{1}{2}p\right)^2=\frac{1}{4}p^2-4$

Taking the square root:

$\displaystyle x+\frac{1}{2}p=\pm\sqrt{\frac{1}{4}p^2-4}$

Solving for x:

$\displaystyle x=-\frac{1}{2}p\pm\sqrt{\frac{1}{4}p^2-4}$

b) If the equation won't have real roots, then the radicand should be negative:

$\displaystyle \frac{1}{4}p^2-4$

Factoring:

$\displaystyle \left(\frac{1}{2}p-2\right)\left(\frac{1}{2}p+2\right)$

The zeros of the left-side polynomial are:

$\displaystyle \frac{1}{2}p-2=0$

$\displaystyle \frac{1}{2}p=2$

p = 4

$\displaystyle \frac{1}{2}p+2=0$

$\displaystyle \frac{1}{2}p=-2$

p = -4

The inequality:

$\displaystyle \left(\frac{1}{2}p-2\right)\left(\frac{1}{2}p+2\right)$

Is satisfied for values of p in the interval (-4,4)

5 0

3 years ago

Fill in the blanks

Tema [17]

1.squares
2.rectangles
3.squares are parralelograms
4.polygons

7 0

3 years ago

a buret is a tool designed to transfer precise amounts of liquid .A buret initially contains 70.00 millimeters of a solution and

katrin [286]

Let:

Vbu= Volume of the buret

Vbk= Volume of the beaker

A buret initially contains 70.00 millimeters of a solution and a beaker initially contains 20.00 ml of the solution the buret drips solution into the Beaker. each drip contains 0.05 mL of solution, therefore:

x = Number of drips

a = volume of each drip

$\begin{gathered} Vbu=70-ax \\ Vbk=20+ax \\ \text{where:} \\ a=0.05 \\ Vbu=70-0.05x \\ Vbk=20+0.05x \end{gathered}$

after how many drips will the volume of the solution in the buret and beaker be equal ? Vbu = Vbk:

$\begin{gathered} Vbu=Vbk \\ 70-0.05x=20+0.05x \\ \text{Solve for x:} \\ 0.1x=70-20 \\ 0.1x=50 \\ x=\frac{50}{0.1} \\ x=500 \end{gathered}$

5 0

1 year ago

I'm kind of unsure of my answer, please help me. I appreciate your answers.

pentagon [3]

Answer:

We can call the numbers x and x + 1. Then we can write:

x + x + 1 = 113 which simplifies to 2x + 1 = 113.

5 0

3 years ago

Read 2 more answers

Why are there three partial products in 481x352 and only two partial products in 481x302?

Sindrei [870]

352 can be separated by 300, 50, and 2. These numbers can produce 3 partial products.

302 can only produce 2 partial products because the only numbers to consider are 300 and 2. 0 is not considered because any number multiplied by 0 is equal to 0.
481 ( 400 + 80 + 1)
 x352 (300 + 50 + 2)
2 (2 x 1)
160 (2 x 80)
800 (2 x 400) Partial product 1 = 962
50 (50 x 1)
4000 (50 x 80)
20000 (50 x 400) Partial product 2 = 24,050
300 (300 x 1)
24000 (300 x 80)
 120000 (300 x 400) Partial product 3 = 144,300
169,312

3 0

3 years ago