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steposvetlana [31]
3 years ago
13

Tasha can spend more than $40 on new boots. Write and graph an inequality to describe how much she can spend.

Mathematics
1 answer:
Svetllana [295]3 years ago
8 0

Answer:

x> \$40

The graph in the attached figure

Step-by-step explanation:

Let

x-----> money that Tasha can spend on new boots

we know that

The inequality that represent the situation is equal to

x> \$40

The solution is the interval--------> (40,∞)

All real numbers greater than 40

In a number line the solution is the shaded area to the right of the dashed line at 40 (open circle)

see the attached figure

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Help me find answer to this
Debora [2.8K]

Answer:

\angle B=41.81^\circ

Step-by-step explanation:

<u>Trigonometric Ratios</u>

The ratios of the sides of a right triangle are called trigonometric ratios.

The longest side of the triangle is called the hypotenuse and the other two sides are the legs.

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\displaystyle \sin B=\frac{\text{opposite leg}}{\text{hypotenuse}}

Here, the opposite leg to angle B is the side CA=2, and the hypotenuse is the side AB=3, thus:

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3 years ago
Find the distance from the point (1,4) to the line y = 1/3x - 3
Troyanec [42]

Answer:

Step-by-step explanation:

If I'm not mistaken, and I very well could be, this is a calculus problem(?). In order to find the distance without calculus you'd need a point on the given line to use to find the distance in the distance formula. But you don't have a point on the given line, so we can find the shortest distance between the point (1, 4) and the given line using the derivative of the polynomial formed when using the distance formula.

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} and we have the x and y for x2 (or x1...it doesn't matter which you choose to fill in):

d=\sqrt{(1-x)^2+(4-y)^2}

but what we find is that we have too many unknowns here, namely, the distance, the x coordinate, and the y coordinate. So we can replace the y coordinate with what y is equal to in terms of the linear equation:

d=\sqrt{(1-x)^2+(4-\frac{1}{3}x-3)^2 } and simplify:

d=\sqrt{(1-x)^2+(7-\frac{1}{3}x)^2 }

. No we'll expand each binomial by squaring:

d=\sqrt{(1-2x+x^2)+(49-\frac{14}{3}x+\frac{1}{9}x^2)  }

.  Combining like terms gives us

d=\sqrt{\frac{10}{9}x^2-\frac{20}{3}x+50  }

The distance between the point (1, 4) and the given line will be at a minimum when the polynomial above is at a minimum. We find the value of x for which the polynomial is at a minimum by finding its derivative, setting the derivative equal to 0, and then solving for x. The derivative of the polynomial is

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Solving for x gives us

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d=\sqrt{(\frac{10}{9})(3)^2-(\frac{20}{3})(3)+50   }

which simplifies down, finally, to

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