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svetoff [14.1K]
3 years ago
10

Write an expression for: the product of 4 and a number. Use x as the variable. Who ever gets it correct first gets brainliest.

Mathematics
1 answer:
maksim [4K]3 years ago
7 0
X(4*1) because I’m just doing what I’m told.
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LeAnn is playing a math game. She chooses three cards. The value of each of her cards is shown.
professor190 [17]

Answer:

-14

Step-by-step explanation:

The sum (addition) of the three numbers would be -14 because -12+-5 is -17. Then, -17+3 would be -14.

6 0
2 years ago
A line has a slope of and a y-intercept of –2.<br><br> What is the x-intercept of the line?
irakobra [83]

Answer:The equation of the line that has a slope of -1/2 and a y-intercept of -2 is: y=-1/2x-2. All you really have to do here is replace m with -1/2, which is the slope you gave, and replace b with -2, the y-intercept you gave, in the equation y=mx+b. 

If this helped, please mark me brainliest, thank you and if you need more help, feel free to message me .

Step-by-step explanation:

8 0
3 years ago
There are (7^13)^3 ⋅ 7^0 strawberries in a field. What is the total number of strawberries in the field?
Dahasolnce [82]
Laws of exponents:
(a^x)^y=a^(xy)
a^0=1  (a&ne;0)

Therefore
(7^13)^3.7^0
=(7^(13*3)). 1
=7^39
5 0
3 years ago
Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular
kozerog [31]
Hello : 
let A(0,3,2) and (Δ) this line , v vector   parallel to (<span>Δ).
M</span>∈ (Δ) : vector (AM) = t v..... t ∈ R

1 )    (Δ)  parallel to the plane x + y + z = 5 : let  : n an vector <span>perpendicular 
to the plane : n </span>⊥ v   ....   n(1,1,1) so : n.v =0  means : n.vector (AM) = 0
(1)(x)+(1)(y-3)+(1)(z -2) =0      ( vector (AM) = ( x, y -3 , z-2 )
x+y+z - 5=0 ...(1)

2)  (Δ) perpendicular to the line (Δ') : x = 1+t  , y = 3 - t , z = 2t :
vector (u) ⊥ v     .... vector(u) parallel to (Δ')  and vector(u) = (1 , -1 ,1)
vector (u) ⊥ vector (AM) means : 
(1)(x)+(-1)(y-3)+(2)(z -2) =0
x - y+2z - 1 = 0 ...(2)
so the system : 
x+y+z - 5=0 ...(1)
x - y+2z - 1 = 0 ...(2)
 (1)+(2) :   2x+3z - 6 =0
x = 3 - (3/2)z
subsct in (1) :    3 - (3/2)z  +y +z - 5 =0
y = 1/2z +2
let : z=t     
an parametric equations for the line (Δ) is :  x = 3 - (3/2)t
                                                                      y = (1/2)t +2
                                                                      z=t

verifiy : 
1) (Δ)  parallel to the plane x + y + z = 5 : 
(-3/2 , 1/2 ,1) <span>perpendicular to (1,1,1)
</span>because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 0
2) (Δ) perpendicular to the line (Δ') :
 (-3/2 , 1/2 ,1) perpendicular to (1,-1,2)
because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0
A(0, 3, 2)∈(Δ) : 
0 = 3-(3/2)t
3 = (1/2)t+2
2 =t
same :  t = 2

3 0
3 years ago
Which graphs show a proportional relationship? Select all that apply.
BlackZzzverrR [31]

Answer:

A, D

Step-by-step explanation:

The graph of a proportional relation is a straight line that passes through the origin.

Answer: A, D

6 0
3 years ago
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