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olga_2 [115]
3 years ago
13

Solve the equation. Check for extraneous solutions please help!!!

Mathematics
1 answer:
lozanna [386]3 years ago
4 0
!z!=a→z=a or z=-a; z=4x+3, a=9+2x

!4x+3!=9+2x
1) 4x+3=9+2x
Solving for x:
4x+3-3-2x=9+2x-3-2x
2x=6
2x/2=6/2
x=3

Checking for extraneous solution:
!4x+3!=9+2x
x=3→!4(3)+3!=9+2(3)
!12+3!=9+6
!15!=15
15=15 Ok, then x=3 is not a extraneous solution 

2) 4x+3=-(9+2x)
Solving for x:
4x+3=-9-2x
4x+3-3+2x=-9-2x-3+2x
6x=-12
6x/6=-12/6
x=-2

Checking for extraneous solution:
!4x+3!=9+2x
x=-2→!4(-2)+3!=9+2(-2)
!-8+3!=9-4
!-5!=5
5=5 Ok, then x=-2 is not a extraneous solution

Answer:
x = -2 or 3 
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7 0
3 years ago
What is 270° converted to radians?<br><br> A.) pi/6<br> B.) 3/2<br> C.) 3pi/2<br> D.) 3
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3 0
3 years ago
PLEASE ANSWER ALL
mars1129 [50]

Step-by-step explanation:

The axis of symmetry: is the line that makes the parabola split in exactly half and lines up with the vertex. For that parabola x=1 is the line of symetry.

The vertex is where the minimum of the graph is, on this graph you can eyeball it to be (1,-9)

The x-intercept is where y is 0 so that's where the lines intersex with the x-axis. (-2,0) and (4,0)

The y-intercept of the function is where x is 0 and where the parabola intersects with the y-axis. On this graph it would be (0,-8)

Hope that helps :)

4 0
3 years ago
Given: ABCD is a trapezoid, AC ⊥ CD AB = CD, AC=the square root of 75 , AB = 5 Find: AABCD
Ugo [173]

In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = \sqrt{75+25} = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that \frac{1}{2} * BF =  \frac{1}{2} * 5 * \sqrt{75} and BF= 0.5\sqrt{75}. Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is  A= BF *  \frac{AB+CD}{2} = 0.5  \sqrt{75}  * \frac{5+10}{2} = 18.75 \sqrt{3}   cm^{2}

3 0
3 years ago
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