In the xy-plane, if (-1, 0) is a solution to the system of inequalities y < x + c and y < -x - d, which of the following must be true about c and d? (10 POINTS + BRAINLIEST ANSWER!!)

A. c = d

B. c = -d

C. c < d

D. d < c

Let xbe a positive integer number. Then x, x+1 and x+2 are three positive consecutive integers (the first one is x, the second is x+1 and the third is x+2).

The product of the second integer and the third integer is (x+1)·(x+2) and is equal to 72. So you have the equation

[tex](x+1)\cdot (x+2)=72.[/tex]

Solve it:

[tex]x^2+2x+x+2=72,\\ \\x^2+3x+2-72=0,\\ \\x^2+3x-70=0,\\ \\D=3^2-4\cdot (-70)=9+280=289,\\ \\\sqrt{D}17,\\ \\x_1=\dfrac{-3-17}{2}=-10,\ x_2=\dfrac{-3+17}{2}=7.[/tex]

Solution [tex]x_1=-10[/tex] is extra because [tex]x_1[/tex] is negative.

Answer: three positive consecutive integers are 7, 8 and 9.

Answer:

ROI = 150

Step-by-step explanation:

ROI = (25,000-10,000)/10,000 ×100=150

Final answer:

To determine the diameter of a circle with a circumference of 12.56 feet using π as 3.14, divide the circumference by π resulting in a diameter of 4 feet.

Explanation:

The student has asked to find the length of the diameter of a circle if the circumference is 12.56 ft using 3.14 for π (Pi). The formula for the circumference C of a circle is C = π * d, where d is the diameter of the circle. To find the diameter, we rearrange the formula to d = C / π. Substituting the given values, we get:

d = 12.56 ft / 3.14

After dividing the circumference by pi, we find that:

d = 4 ft

Therefore, the length of the diameter is 4 feet.

The optimum angle at which a projectile should be launched to cover maximum distance is 45°. This angle gives the projectile the greatest range by maximizing the time of flight.

The optimum angle at which a projectile should be launched in order to cover the maximum distance is 45°. This launch angle gives the projectile the greatest range.

To understand why this angle is optimal, we can consider that the motion of a projectile can be divided into two independent components: horizontal and vertical. The range of a projectile is maximized when the time it spends in the air is maximized. At a launch angle of 45°, the horizontal and vertical components of the projectile's velocity are equal. This means that the time of flight is maximized, resulting in the maximum range.

For example, if a projectile is launched at a shallower angle, such as 30°, the vertical component of the velocity is decreased, reducing the time of flight and therefore the range. Similarly, if the projectile is launched at a steeper angle, such as 60°, the horizontal component of the velocity is decreased, again reducing the time of flight and range.

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Look at changes of signs to find this has 1 positive zero, 1 or 3 negative zeros and 0 or 2 non-Real Complex zeros.

Then do some sums...

f(x)=−3x4−5x3−x2−8x+4

Since there is one change of sign, f(x) has one positive zero.

f(−x)=−3x4+5x3−x2+8x+4

Since there are three changes of sign f(x) has between 1 and 3 negative zeros.

Since f(x) has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so f(x) has exactly 1 or 3 negative zeros counting multiplicity, and 0 or 2 non-Real Complex zeros.

f'(x)=−12x3−15x2−2x−8

Newton's method can be used to find approximate solutions.

Pick an initial approximation a0 .

Iterate using the formula:

ai+1=ai−f(ai)f'(ai)

Putting this into a spreadsheet and starting with a0=1 and a0=−2 , we find the following approximations within a few steps:

We can then divide f(x) by (x−0.42) and (x+2.195) to get an approximate quadratic −3x2+0.325x−4.343 as follows:

Notice the remainder 0.013 of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.

Check the discriminant of the approximate quotient polynomial:

Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly 2 non-Real Complex zeros, 1 positive zero and 1 negative one.

Answer:

Positive: 2 or 0; negative: 2 or 0; complex: 4 or 2 or 0

To find the length marked x, establish the ratio 0.5 inch/20 miles = 8 inches/x miles, cross-multiply, and solve for x to get x = 320 miles, making sure to round only after the final calculation step.

To solve for the length labeled x, you would first need to set up the correct ratio. Given that the scale length is 8 inches and the corresponding actual length is unknown, the initial ratio would be 0.5 inch/20 miles = 8 inches/x miles. You can solve this proportion by cross-multiplication.

Following these steps:

Therefore, the unknown length x is 320 miles. Remember to always perform rounding off at the final step of your calculation to ensure accuracy.

The length of side [tex]\( x \)[/tex] is approximately 13.9 units when rounded to the nearest tenth.

To find the length of side [tex]\( x \)[/tex], we utilize the tangent function because it relates the opposite side to the adjacent side in a right-angled triangle. The tangent of [tex]\( 41^\circ \)[/tex] equals the ratio of side [tex]\( x \)[/tex] (the side opposite to [tex]\( 41^\circ \))[/tex] to 16 (the side adjacent to [tex]\( 41^\circ \))[/tex]:

[tex]\[ \tan(41^\circ) = \frac{x}{16} \][/tex]

To solve for [tex]\( x \)[/tex], we multiply both sides by 16:

[tex]\[ x = 16 \times \tan(41^\circ) \][/tex]

[tex]x=13.9[/tex]

The length of side [tex]\( x \)[/tex] is approximately 13.9 units when rounded to the nearest tenth.

The correct solution of the quadratic equation 3x^2 - x = 10 is found using the quadratic formula, giving two solutions x = -5/3 and x = 2.

The equation given is 3x2 − x = 10, which is a quadratic equation that needs to be rearranged to the standard form ax2 + bx + c = 0.

First, we subtract 10 from both sides of the equation, resulting in 3x2 − x − 10 = 0. Now, we can use the quadratic formula:

x = −b ± √(b2 - 4ac) / (2a)

Substituting the values a = 3, b = −1, and c = −10 into the formula yields:

x = −(−1) ± √((−1)2 - 4 × 3 × (−10)) / (2 × 3)

This simplifies to:

x = 1 ± √(1 + 120) / 6

Which further simplifies to:

x = 1 ± √121 / 6

And results in:

x = 1 ± 11 / 6

Therefore, the two solutions are:

x = 2 (1 + 11 / 6)

x = − five over three (1 − 11 / 6)

Hence, the correct answer is x = − five over three and x = 2.

The quadratic equation 3x^2 - x = 10 is solved via the quadratic formula, resulting in two solutions: x = 2 and x = -5/3 (or -1.67 when expressed as a decimal).

To solve the quadratic equation 3x2 − x = 10, we first need to set it to zero by subtracting 10 from both sides, yielding 3x2 − x − 10 = 0. This is a standard form quadratic equation, ax2 + bx + c = 0, where a = 3, b = -1, and c = -10. We can solve for x by using the quadratic formula, x = ∛ ( b2 - 4ac ) / (2a).

Plugging in our values, we get:

x = (-(-1) ± √ ( (-1)2 - 4(3)(-10) ) ) / (2(3))
x = (1 ± √ ( 1 + 120) ) / 6
x = (1 ± √ (121) ) / 6
x = (1 ± 11) / 6

Thus:

Therefore, the solutions are x = 2 and x = -5/3.

The brightness of a light bulb in watts is considered continuous quantitative data as it represents measurable amounts that can vary and take on any value within a range.

The brightness of a light bulb, measured in watts, represents the power it uses and is an example of quantitative data. Since wattage can take on any positive value and is not restricted to whole numbers (for example, a bulb can be rated at 60.5 watts), it is considered to be continuous quantitative data. Unlike discrete data, which involve counts that must be whole numbers, continuous data can include any value within a range. Hence, when comparing the brightness of light bulbs by their wattage, one is dealing with continuous quantitative data.