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allochka39001 [22]
3 years ago
7

Please help me please

Mathematics
2 answers:
yaroslaw [1]3 years ago
6 0

Answer:

B.

Step-by-step explanation:

The x coordinates stay the same. If you actually look at the points, the x coordinate stays the same, while the y coordinate is now negative.

dsp733 years ago
4 0

Answer:

B

Step-by-step explanation:

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Consider the quadratic equation. x^2=4x-5. How many solutions does the equation have?
Vesna [10]

x^2=4x-5
subtract 4x from both sides
x^2-4x=-5
add 5 to both sides
x^2-4x+5=0

input into quadratic formula which is x=\frac{-b+ \sqrt{b^2-4ac} }{2a} or \frac{-b- \sqrt{b^2-4ac} }{2a}

si ax^2+bx+c
so a=1
b=-4
c=5
input
\frac{-(-4)+ \sqrt{-4^2-4(1)(5)} }{2(1)}=\frac{4+ \sqrt{16-20} }{2(1)}=\frac{4+ \sqrt{-4} }{2}=\frac{4+ \sqrt{4} times \sqrt{-1} }{2} \frac{4+2 times  \sqrt{-1}  }{2}=  \frac{6 times  \sqrt{-1}  }{2}=3 times  \sqrt{-1} [\tex][\tex]\sqrt{-1} representeds by 'i' so solution is 3i

then if other way around then wyou would do
\frac{-(-4)- \sqrt{-4^2-4(1)(5)} }{2(1)}=\frac{4- \sqrt{16-20} }{2(1)}= \frac{4- \sqrt{-4} }{2} =\frac{4- \sqrt{4} times \sqrt{-1} }{2}= \frac{4-2 times \sqrt{-1} }{2}=\frac{2 \sqrt{-1} }{2}= \sqrt{-1} and [\tex]\sqrt{-1} [/tex] is represented by i


the solution is x=3i or i (i=\sqrt{-1})
but i is not real, it is imaginary so there are no real solution so the answer is C



3 0
3 years ago
What is the area of a rectangle with vertices (-8, -2), (-3,-2).(-3,-8), and (-8, -B)?
den301095 [7]

Answer:

The area of the rectangle would be 20 square units.

Step-by-step explanation:

The distance between the points (-8, -2) and (-3,-2) is 5. This is the length of the rectangle.  The distance between (-3,-6) and (-3,-2) is 4. This is the width of the rectangle. To find the area, multiply the length and the width to get the 20 square units as the area.

6 0
3 years ago
Read 2 more answers
ANSWER ANY PART YOU KNOW!! (you don't have to answer all)
jolli1 [7]

Answer:

a)2*(15-s) mi

b)30mi

c) 30 mi

Step-by-step explanation:

6 0
3 years ago
Solve the quadratic equation using the quadratic formula: 3x^2 - 6x = 5
Fantom [35]

Answer:

x_{1} = 1+\frac{2\sqrt{6}}{3} \\\\x_{2} = 1-\frac{2\sqrt{6}}{3}

Step-by-step explanation:

3x^2 - 6x - 5=0

quadratic formula:

x=\frac{-b+-\sqrt{b^{2}-4*a*c}}{2*a}

a=3     b=-6     c=-5

x=\frac{-(-6)+-\sqrt{(-6)^{2}-4*3*-5}}{2*3}

x=\frac{6+-\sqrt{36+60}}{6} \\\\x=\frac{6+-\sqrt{96}}{6} \\\\x=\frac{6+-\sqrt{2^{2}*2^{2}*2*3 }}{6} \\\\x=\frac{6+-4\sqrt{6}}{6} \\

so we have:

x_{1} = 1+\frac{2\sqrt{6}}{3} \\\\x_{2} = 1-\frac{2\sqrt{6}}{3}

8 0
3 years ago
Please help on this pythagorean question, find x.
svp [43]

Answer:

x = 4

Step-by-step explanation:

Since the triangle is right use Pythagoras' identity to solve for x

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

(x + 3)² + (4(x + 2))² = 25² ← expand parenthesis on left side

x² + 6x + 9 + 16(x+ 2)² = 625

x² + 6x + 9 + 16(x² + 4x + 4) = 625

x² + 6x + 9 + 16x² + 64x + 64 = 625 ← simplify left side

17x² + 70x + 73 = 625 ( subtract 625 from both sides )

17x² + 70x - 552 = 0 ← in standard form

with a = 17, b = 70, c = - 552

Using the quadratic formula to solve for x

x = ( - 70 ± \sqrt{70^2-(4(17)(-552)} ) / 34

  = ( - 70 ± \sqrt{4900+37536} ) / 34

  = - 70 ± \sqrt{42436} ) / 34

  = - 70 ± 206 ) / 34

x = \frac{-70-206}{34} = - 8.1176....

or x = \frac{-70+206}{34} = 4

However, x > 0 ⇒ x = 4

Hence

x + 3 = 4 + 3 = 7 and

4(4 + 2) = 24

The triangle is a 7- 24- 25 right triangle

7 0
3 years ago
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