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AfilCa [17]
3 years ago
13

Calculeaza2+(8-6)=(9-7)+5=10-(1+8)=(3+7)-9=(1+2+3)-5=10-(2+3+4)=​

Mathematics
1 answer:
Alex787 [66]3 years ago
6 0

Answer:

1. 2+(8-6)= 4

2+2=4

2. (9-7)+5= 7

2+5=7

3. 10-(1+8)= 1

10-9=1

4. (3+7)-9= 1

10-9=1

5. (1+2+3)-5= 1

6-5=1

6. 10-(2+3+4)=​ 1

10-9=1

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TV=3x-24 and VX=2x+1. What is the value of VX?<br> 1. 25<br> 2. 99<br> 3. 5<br> 4. 51
Maslowich
ΔTXV is an <span>isosceles triangle because ∡T = ∡X therefore TV = VX.

3x - 24 = 2x + 1    |subtract 2x from both sides

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3 years ago
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Which of the following is a solution set to the equation x^2-10x+16=0
almond37 [142]

Answer:

x=2. or. x=8

Step-by-step explanation:

x^2-10x+16=0. s=-10

x^2-8x-2x+16=0. p=16

x(x-8)-2(x-8)=0. [-8,-2]

(x-2)(x-8)

x=2. or. x=8

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3 years ago
The value of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B0%7D%7B%28m%2Bn%29%7D" id="TexFormula1" title="\frac{0}{(m+n)}" alt="\f
Sever21 [200]

Answer:

two in the middle

Step-by-step explanation:

6 0
3 years ago
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Please help i need to get this done asap !
xz_007 [3.2K]

Answer:

Step-by-step explanation:

Equation of line k is   y = (-4/7)x + 1

point (-5,-10); m = -4/7

Line L is perpendicular to k so the perpendicular slope will be 7/4

Equation of line L is  (y - y1) = m(x-x1)

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3 0
3 years ago
Find all relative extrema of the function. Use the Second Derivative Test where applicable. (If an answer does not exist, enter
Lapatulllka [165]

Answer:

Relative minimum: \left(-\frac{5}{2}, -\frac{33}{4}\right), Relative maximum: DNE

Step-by-step explanation:

First, we obtain the First and Second Derivatives of the polynomic function:

First Derivative

f'(x) = 2\cdot x + 5 (1)

Second Derivative

f''(x) = 2 (2)

Now, we proceed with the First Derivative Test on (1):

2\cdot x + 5 = 0

x = -\frac{5}{2}

The critical point is -\frac{5}{2}.

As the second derivative is a constant function, we know that critical point leads to a minimum by Second Derivative Test, since f\left(-\frac{5}{2}\right) > 0.

Lastly, we find the remaining component associated with the critical point by direct evaluation of the function:

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There are relative maxima.

6 0
3 years ago
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