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SOVA2 [1]
3 years ago
14

Please help ?!!! Pleaee

Mathematics
1 answer:
Taya2010 [7]3 years ago
6 0
31 is the answer; this is because to get to the place-holder number “0” you have to add 19 to -19, then after this, you add 12, making the total 31.
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Descompuneti in factori. x^2+4x-5=?
Lapatulllka [165]
<span>Assume that,
   x^2+4x-5 = 0 .......(1)
       Then, x^2+4x-5 = 0 x^2+5x-1x-5 =0
   x(x+5)-1(x+5) = 0
    (x+5) (x-1) = 0
       We get x=-5 and x=1
       Sub x=-5 in equ (1)
   (-5)^2+4(5)-5 = 0
     -25+20-5 = 0
    -25+25= 0
    0 = 0
       Sub x=1 in equ (1) (1)^2+4(1)-5 = 0
    1+4-5 = 0 5-5 = 0
     0 = 0
       Therefore x value is -5 and 1</span>
3 0
3 years ago
-5(2 × -6) inverse operation
Gekata [30.6K]

Answer:

60

Step-by-step explanation:

-5(2×-6)

-5(-12)

60

It should be correct

5 0
2 years ago
Which sets of values can be the measures of the exterior angles of a triangle?
romanna [79]
The 3 angles must add up to 360 degrees

Only 1 and 4  fit the bill.
5 0
3 years ago
A projectile is thrown upward off a 130ft cliff at a velocity of 15ft/second. How long before it reaches a height of 25ft?
zhuklara [117]

Answer:

The time taken for the projectile to reach the given height is 1.2 s.

Step-by-step explanation:

Given;

height of travel, h = 25 ft

initial velocity of the projectile, u = 15 ft/s

The time taken for the projectile to travel a height of 25 ft is given by the following kinematic equation;

h = ut + ¹/₂gt²

25 = 15t + ¹/₂(9.8)t²

25 = 15t + 4.9t²

4.9t² + 15t - 25 = 0

solving the quadratic equation, we will have the following solution of t;

t = 1.2 s

Therefore, the time taken for the projectile to reach the given height is 1.2 s.

5 0
2 years ago
Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows
never [62]

Answer:

a.  For the first lake;

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m·t/200 - m·t²/80000

b. t ≈ 1.00505 hours

c. 200 hours

Step-by-step explanation:

The flow rate of water in and out of the lakes = 500 liters/hour

The volume of water in the first lake = 100 thousand liters

The volume of water in the second lake = 200 thousand liters

The mass of toxic substances that entered into the first lake =  500 kg

The concentration of toxic substance in the first lake = m₁/(100000)

Therefore, we have;

The quantity of fresh water supplied at t hours = 500 × t

The change

The change in the mass of the toxic substance with time is given as follows

dm/dt = (m - m/100000 × 500 × t)/100000

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m/100000 × 500 × t - (m/100000 × 500 × t)/200000 × 500 × t

Using an online tool, we have;

c = m·t/200 - m·t²/80000

b. When c < 0.001 kg per liter, we have m < 0.001 × 100000, which gives m < 100

Substituting gives;

0.001 = (100·t - 0.005·100·t²)/100000, solving with an online tool, gives;

t ≈ 1.00505 hours

c. For maximum concentration, we have;

c = m·t/200 - m·t²/80000

m/200000 = m·t/200 - m·t²/80000

1/200000 = t/200 - t²/80000

dc/dt = d(t/200 - t²/80000)/dt = 0

Solving with an online tool gives t = 200 hours

6 0
2 years ago
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