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Sergeu [11.5K]
3 years ago
12

Four less than five times a number is equal to 11

Mathematics
2 answers:
natka813 [3]3 years ago
7 0

x is the number

5x - 4 = 11

tankabanditka [31]3 years ago
5 0
5-4*N=11 

this is how you would write it 

hope this helps
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Someone help me with this question please.
Usimov [2.4K]

Answer:

no

Step-by-step explanation:

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6 0
3 years ago
If 4x^2 - 100 = 0, the roots of the equation are
ololo11 [35]
4x^2-100=0 \ \ \ \ \ \ |+100 \\
4x^2=100 \ \ \ \ \ \ \ \ \ \ \  |\div 4 \\
x^2=25 \\
x=\pm \sqrt{25} \\
x=\pm 5 \\
x=-5 \hbox{ or } x=5

The answer is (3) -5 and 5.
5 0
3 years ago
Chase consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Chase's body
PIT_PIT [208]

Answer:

(a) The 5-hour decay factor is 0.5042.

(b) The 1-hour decay factor is 0.8720.

(c) The amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

Step-by-step explanation:

The amount of caffeine in Chase's body decreases exponentially.

The 10-hour decay factor for the number of mg of caffeine is 0.2542.

The 1-hour decay factor is:

1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720

(a)

Compute the 5-hour decay factor as follows:

5-hour\ decay\ factor=(0.8720)^{5}\\=0.504176\\\approx0.5042

Thus, the 5-hour decay factor is 0.5042.

(b)

The 1-hour decay factor is:

1-hour\ decay\ factor=(0.2542)^{1/10}=0.8720

Thus, the 1-hour decay factor is 0.8720.

(c)

The equation to compute the amount of caffeine in Chase's body is:

A = Initial amount × (0.8720)<em>ⁿ</em>

It is provided that initially Chase had 171 mg of caffeine, 1.39 hours after consuming the drink.

Compute the amount of caffeine in Chase's body 2.39 hours after consuming the drink as follows:

A = Initial\ amount \times (0.8720)^{2.39} \\=[Initial\ amount \times (0.8720)^{1.39}] \times(0.8720)\\=171\times 0.8720\\=149.112

Thus, the amount of caffeine in Chase's body 2.39 hours after consuming the drink is 149.112 mg.

4 0
3 years ago
If triangle HIJ is dilated about the center of the triangle to create triangle H'I'J', dilated line A'B' will
nevsk [136]
Pls. see attachment.

7 0
3 years ago
Read 2 more answers
At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to
In-s [12.5K]

The given equation of the ellipse is x^2 + y^2 = 2 x + 2 y

At tangent line, the point is horizontal with the x-axis therefore slope = dy / dx = 0

<span>So we have to take the 1st derivative of the equation then equate dy / dx to zero.</span>

x^2 + y^2 = 2 x + 2 y

x^2 – 2 x = 2 y – y^2

(2x – 2) dx = (2 – 2y) dy

(2x – 2) / (2 – 2y) = 0

2x – 2 = 0

x = 1

 

To find for y, we go back to the original equation then substitute the value of x.

x^2 + y^2 = 2 x + 2 y

1^2 + y^2 = 2 * 1 + 2 y

y^2 – 2y + 1 – 2 = 0

y^2 – 2y – 1 = 0

Finding the roots using the quadratic formula:

y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1

y = 1 ± 2.828

y = -1.828 , 3.828

 

<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).</span>

3 0
3 years ago
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