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Daniel [21]
3 years ago
11

Solve for x in the equation x2 - 4x-9= 29.

Mathematics
1 answer:
alukav5142 [94]3 years ago
7 0

Answer:

I think the first one

Step-by-step explanation:

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Explain the difference between the theoretical probability and finish odds
stepladder [879]

Answer:

Theoretical probability is what that you expect to happen to you while experimental is what actually happen when you try it out.

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3 years ago
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A box has a perimeter of 27 1/2 ft. if the lenght is 9 ft. what is the area?
gavmur [86]

Answer:

Perimeter = (Length x 2) + (Width x 2)

27.5 = (9 x 2) + (Width x 2)

27.5 = 18 + (Width x 2)

9.5 = (Width x 2)

4.75 = Width

check:

27.5 = (9 x 2) + (4.75 x 2)

27.5 = 18 + 9.5

27.5 = 27.5

Since the width is 4.75 and the length is 9, we can multiply the two to get the area. 4.75 x 9 = 42.75 sq. ft.

8 0
3 years ago
What must be the value of x so that lines a and b are parallel lines cut by transversal f?
ExtremeBDS [4]

we know that

if the lines a and b are parallel lines cut by transversal f

then

(6x-36)=96 --------> by alternate exterior angles

solve for x

(6x-36)=96\\ 6x=96+36\\6x=132\\x=(132/6)\\x=22\ degrees

therefore

<u>the answer is</u>

the value of x is 22\ degrees

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3 years ago
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Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 51 km/h. After five hou
bija089 [108]
It's not obvious here, but you're being asked to find a linear equation for the velocity of the car, given two points on the line that represents this velocity.

Find the slope of the line segment that connects the points (3 hr, 51 km/hr) and (5 hr, 59 km/hr).  Graph this line.  Where does this line intersect the y-axis?  Find the y-value; it's your "y-intercept," b.

Now write the equation:  velocity = (slope of line)*t + b

The units of measurement of "slope of line" must be "km per hour squared," and those of the "y-intercept" must be "km per hour."

Part B:  Start with the y-intercept (calculated above).  Plot it on the vertical axis of your graph.  Now label the horizontal axis in hours:  {0, 1, 2, 3, 4, 5, 6}.  Draw a vertical line through t=6 hours.  It will intercept both the horiz. axis and the sloping line representing the velocity as a function of time.  Show only the part of the graph that extends from t=0 hours to t=6 hours.

6 0
3 years ago
Let V denote the set of ordered triples (x, y, z) and define addition in V as in
icang [17]

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. (u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)

5. (u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,y,az)

b)  6. c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du

9.

c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u

10

1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,0,az)

c) Observe that 1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(0,0,0).

d)  Observe that 1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(2ax,2ay,2az).

8 0
3 years ago
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