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BaLLatris [955]
3 years ago
9

2= 5 j -28 one or two step equation

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

j=6

Step-by-step explanation:

2=5j-28

+28  +28

30=5j

/5    /5

j=6  

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Solve the simultaneous equations
NeTakaya

Answer:

y = 2, x = -1

Step-by-step explanation:

2y - 5x = 9

2y = 9+5x

y = (9+5x) /2

4y + 3x = 5

4((9+5x) /2) + 3x = 5

13x + 18 = 5

13x = -13

x = -1

y = (9+5x) /2

y = (9+5(-1)) /2

y = 2

4 0
2 years ago
Read 2 more answers
A large department store is prepared to buy 3,800 of your tie-dye shower curtains per month for $5 each, but only 3,600 shower c
Ksivusya [100]

Answer:

q= -40p+c

q= -40p+4,000

Step-by-step explanation:

$5 each for 3800 tye-dye shower curtains

$10 each for 3600 tye-dye shower curtains

Slope of the demand line is

3800-3600/5-10

=200/-5

= -40

Demand function is

q= -40p + c

Where c is a constant

For q=3800 and p=$5

Demand function is

q= -40p + c

3800= -40(5)+c

3800= -200+c

c=3800+200

c=4,000

Linear demand function is

q= -40p+4,000

6 0
3 years ago
Given the equation y = 3x − 4, what is the value of y when x = 4?
Marat540 [252]

Answer:

C) 8

Step-by-step explanation:

y=3*4-4

y=12-4

y=8

5 0
3 years ago
6 - 3x = 5x - 10x + 8
Jet001 [13]

Answer:

x=1

Step-by-step explanation:

6-3x=5x-10x+8=

move all terms to the left:

6-3x-(5x-10x+8)=0

add all the numbers together, and all the variables

-3x-(-5x+8)+6=0

get rid of parentheses

-3x+5x-8+6=0

add all the numbers together, and all the variables

2x-2=0

move all terms containing x to the left, all other terms to the right

2x=2

x=2/2

x=1

3 0
3 years ago
At time t ≥ 0, the velocity of a body moving along the s-axis is v = t² -5t +4. When is the body moving backwards
katovenus [111]

Answer:

A

Step-by-step explanation:

The velocity of a moving body is given by the equation:

v=t^2-5t+4 ,\, t\geq0

Is the velocity is <em>positive </em>(v>0), then our object will be moving <em>forwards</em>.

And if the velocity is negative (v<0), then our object will be moving <em>backwards</em>.

We want to find between which interval(s) is the object moving backwards. Hence, the second condition. Therefore:

v

By substitution:

t^2-5t+4

Solve. To do so, we can first solve for <em>t</em> and then test values. By factoring:

(t-4)(t-1)=0

Zero Product Property:

t=1, \text{ and } t=4

Now, by testing values for t<1, 1<t<4, and t>4, we see that:

v(0)=4>0,\, v(2)=-20

So, the (only) interval for which <em>v</em> is <0 is the second interval: 1<t<4.

Hence, our answer is A.

7 0
3 years ago
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