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Naddika [18.5K]
3 years ago
8

5(a−1)−15=3(a+2)+4 a= ?

Mathematics
1 answer:
Alla [95]3 years ago
6 0

Answer:

a = 15

Step-by-step explanation:

screenshots below should explain the answer :)

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Tran is solving the quadratic equation 2x2 – 4x – 3 = 0 by completing the square. His first four steps are shown in the table.
Nady [450]
The following steps of solving for the roots of 2x² - 4x -3 = 0 were retrieved from another source Step 1 2x² - 4x = 3 Step 2 2(x² - 2x) = 3 Step 3 2(x² - 2x + 1) = 3 + 1 Step 4 2(x - 1)² = 4 From this, we can see that on Step 3, Tran made a mistake of adding 1 to 3. As we can see, 2(x² - 2x + 1) = 2x² - 4x + 2. That means, instead of adding 1, it should have been 2. Therefore, the step that Tran first made an error is Step 3<span>.</span>
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3 years ago
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Solve for x.<br><br> 8x^2−3=2
sergeinik [125]

8x^2=2+3

8x^2=5

x^2=5/8

x= sqrt(5/8)

5 0
3 years ago
Find an explicit solution of the given initial-value problem. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0
MissTica

Answer:

a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x*(1 + 4y²) dx = 0

(1 + x⁴) dy  = - x*(1 + 4y²) dx

[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx

∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx

now to solve each integral

I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁

I₂=  ∫[-x/(1 + x⁴)] dx

for u= x² → du=x*dx

I₂=  ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ =  - tan⁻¹ (x²) +C₂

then

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C

for y(x=1) = 0

1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C

since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for  π*N , we will choose for simplicity N=0 . hen an explicit solution would be

1/2 * 0 = - π/4 + C

C= π/4

therefore

1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

5 0
4 years ago
When a sprinkler is installed in the ground, the spray of water goes up and falls in the pattern of a parabola. The height, in i
Westkost [7]

Answer:

(1) 256 inches

(2) 5 feet

(3) 400 inches

(4) 10 feet

Step-by-step explanation:

(1) The function that gives the height in inches of the spray of water at a distance <em>x</em> from the sprinkler head is given as follows;

h(x) = 160·x - 16·x²

At x = 2 feet, we have;

h(2) = 160 × 2 - 16 × 2² = 256

Therefore, the height of the spray water at a horizontal distance of 2 feet from the sprinkler head h(2) = 256 inches

(2) The x-coordinate, x_{max}, of the maximum point of a parabola given in the form, y = a·x² + b·x + c is found using the following formula;

x_{max} = -b/(2·a)

The x-coordinate, x_{max}, of the maximum point of the given equation of the parabola, h(x) = 160·x - 16·x², (a = -16, b = 160) is therefore;

x_{max} = -160/(2 × (-16)) = 5

Therefore, the number of feet along the way, the function will reach maximum height, x_{max} = 5 feet

(3) The function, h(x) = 160·x - 16·x², will reach maximum height, h_{max}, at x = 5, therefore;

h_{max} =  h(5) = 160 × 5 - 16 × 5² = 400

The maximum height of the spray, h_{max} = 400 inches

(4) The water is at ground level where h(x) = 0, therefore;

At ground level, h(x) = 0 = 160·x - 16·x²

160·x - 16·x² = 0

∴ 16·x × (10 - x) = 0

By zero product rule, we 16·x = 0, or (10 - x)  = 0, from which we have;

x = 0, or x = 10

The water is at ground level at x = 0 and x = 10 feet, therefore, the water will hit the ground again (the second time after leaving the sprinkler head at x = 0) at x = 10 feet.

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3 years ago
Write the equation in slope intercept form. Identify the m and b. 3x−3y=−9
lana [24]
X=-3+y is the answer I got
6 0
4 years ago
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