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Westkost [7]
3 years ago
10

Determines whether the line through the points

Mathematics
1 answer:
kotegsom [21]3 years ago
5 0
Find their gradients using the change in y coords divided by the change in x coords. once you have the gradients (or slopes), multiply them by eachother - if the product is (-1) then theyre perpdendicular, if not, they are either parallel or intersect at a point
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Find an integer value of b that makes x^2 + bx - 81 factorable.
Kaylis [27]

Answer:

b = 80

Step-by-step explanation:

Given that,

x² + bx - 81

To find,

value of b

What is factorable?

A polynomial equation with highest degree 2 is if factorable when

  1. we can find two terms which when multiples = -81x²,   <em> ( x² * - 81 = -81x² ) </em>
  2. and when add = bx

possible factors:

  1. -1x * 81x = -81x²   (Accepted)
  2. 1x * -81x= -81x²    (rejected)
  3. 9x * -9x = -81x²    (rejected)
  4. -9x * 9x = -81x²    (rejected)

Only when -1x , 81x is added -1x + 81x = 80x

so, bx = 80x

     b   = 80

<h3>The equation is x² + 80x - 81</h3>

5 0
3 years ago
Write the point slope equation of the line through (7,-2) with slope of 1/2.
ahrayia [7]

Answer:

y = 0.5x - 5.5

Step-by-step explanation:

slope = 0.5, so this is the x coefficient.

y intercept = -2 - (7 x 0.5) = -5.5

4 0
3 years ago
Trapezoid ABCD has an area of 525 cm squared. If height AB = 21 cm and BC = 37 cm, what is the measure of AD?​
Annette [7]

Answer:

AD = 13

Step-by-step explanation:

Area of trapezoid ABCD

=  \frac{1}{2} (AB + AD) \times AB \\  \\  525=  \frac{1}{2} (37+ AD) \times21\\  \\  525 \times 2=  (37+ AD) \times21 \\  \\ 1050 = 21 \times 37 + 21AD \\  \\ 1050  = 777+ 21AD \\  \\  21AD  = 1050 - 777 \\  \\ 21AD  = 273 \\  \\ AD  =  \frac{273}{21}  \\  \\ AD  = 13

8 0
2 years ago
A2 = -8 and a5 = -512<br> Find a10
yawa3891 [41]

Answer:

The value of a₁₀ is -1352

Step-by-step explanation:

a₂ = -8

a₅ = -512

Now,

a₂ = -8 can be written as

a + d = -8 ...(1) and

a₅ = -512 can be written as

a + 4d = -512 ...(2)

Now, from equation (2) we get,

a + 4d = - 512

a + d + 3d = - 512

(-8) + 3d = - 512 (.°. <u>a + d = </u><u>-8</u><u>)</u>

3d = - 512 + 8

3d = - 504

d = - 504 ÷ 3

d = - 168

Now, for the value of a put the value of d = -168 in equation (1)

a + d = -8

a + (-168) = -8

a - 168 = -8

a = 168 - 8

a = 160

Now, For a₁₀

a₁₀ = a + 9d

a₁₀ = 160 + 9(-168)

a₁₀ = 160 - 1512

a₁₀ = -1352

Thus, The value of a₁₀ is -1352

<u>-TheUnknownScientist</u>

8 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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