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2 months ago

Which point is a solution to the inequality shown in this graph?

Mathematics

2 answers:

PIT_PIT [12.4K]2 months ago

6 0

Answer:

A) (0,5)

Step-by-step explanation:

The coordinate (0,5) exists within the shaded blue area

zzz [12.3K]2 months ago

5 0

Answer:

Option A is indeed the right choice, as (0, 5) satisfies the stated inequality.

Step-by-step explanation:

Given

An illustration representing the inequality.

Two points that lie on the line of the inequality are (-3, -6) and (3, 2).

To find: Identify which of the specified points fulfills the inequality.

It is evident from the graph that it depicts a strict inequality since the line indicated is dotted rather than solid.

Thus, points (-3, -6) and (3, 2) qualify as solutions to the inequality.

The point (5, 2) does not qualify since it is situated outside the shaded area.

This leads to the conclusion that Option A is correct in stating that (0, 5) is indeed a solution to the inequality, as it is found within the shaded area.

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1 product = $65.00
20 products = 65 x 20 = $1300

Sales Tax is 3.5% of $1300 = 0.035 x $1300 = $45.50

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8 0

2 months ago

Read 2 more answers

A waterbed filled with water has the dimensions 8ft by 7 ft by .75 ft taking the density of water to be 1.00/gm cubed how many k

Svet_ta [12734]

Density can be defined as:

$D = \frac{m}{V}$

Where:

m is mass

V stands for volume

To isolate the mass, we derive:

$m = DV$

Volume can be expressed as:

$V = (8) * (7) * (0.75)$

$V = 42ft ^ 3$

We then use the following conversion:

$1foot = 0.3048m$

After applying the conversion, we arrive at:

$V = 42 * (0.3048) ^ 3$

$V = 1.19m ^ 3$

Additionally, we have these conversions:

$1m = 100cm$

$1Kg = 1000g$

When we apply these conversions for density, we conclude:

$D = (1\frac{g}{cm^3})((\frac{100}{1})^3\frac{cm^3}{1m^3})(\frac{1}{1000}\frac{Kg}{g})=1000\frac{Kg}{m^3}$

Finally, the mass of the water required is:

$m = (1000) * (1.19)$

$m = 1190$

Answer:

A total of 1190 kilograms of water is needed to fill the waterbed.

7 0

2 months ago

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin

Zina [12379]

Answer:

(a) $\frac{dy}{dt}=-3\frac{3}{4}$

(b) $\frac{dx}{dt}=3\frac{3}{4}$

Step-by-step explanation:

$x^{2} +y^{2}=25$

Calculate $\frac{d}{dt}$ for each term.

$\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\$

For Question a

$2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

With x = 3, y = 4, and dx/dt = 5.

$\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\ \\\frac{dy}{dt}=-3\frac{3}{4}$

For Question b

$2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$

Given x = 4, y = 3, and dx/dt = -5.

$\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\ \\\frac{dx}{dt}=3\frac{3}{4}$

5 0

2 months ago

The area of a square picture frame is 55 square inches. Find the length of one side of the frame. Explain to the nearest whole i

Inessa [12570]

We know that the area of a square picture frame measures 55 square inches. We need to calculate the length of one side of the frame. Let's denote the side length as x inches. The formula for the area of a square is x squared, and we can conclude that the area equals 55. Taking the square root of both sides, we see that 55 is nearer to 49, so x is approximately 7 inches.

7 0

2 months ago