A flower bed has the shape of a rectangle 18 feet long and 12 feet wide. What is its area in square yards?

A person standing 154 feet from the base of a church observed the angle of elevation to the church's steeple to be 25∘. How tall

Nookie1986 [14]

Answer:

71.764 feet

Step-by-step explanation:

See attached the rough sketch for your reference

Step one;

Given data

The angle of elevation= 25°

From the triangular diagram, we can see that the height of the church is the opposite, while the distance of the person from the church is the adjacent

Step two:

Applying SOH CAH TOA

tan∅= opp/adj

tan 25= opp/154

0.466= opp/154

cross multiply

0.466*154= opp

opp= height of church = 71.764 feet

4 0

3 years ago

Solve. −43x+16<79 Drag and drop a number or symbol into each box to show the solution.

Debora [2.8K]

The answer is the third one

7 0

3 years ago

Read 2 more answers

The length plus the width of a rectangle is 10. Let x represent the length. The equation for the area y of the

yuradex [85]

Answer:

length of rectangle = 5

width of rectangle = 5

Area of rectangle = 25

Step-by-step explanation:

Since the length of the rectangle is "x", and the value of the area is given by the product of the length "x" times the width "10-x", indeed, the area "y" of the rectangle is given by the equation:

$y=x\,(10-x)=10x-x^2$

Now, they tell us that the area of the rectangle is such that coincides with the maximum (vertex) of the parabola this quadratic expression represents. So in order to find the dimensions of the rectangle and therefore its area, we find the x-coordinate for the vertex, and from it, the y-coordinate of the vertex, which is the rectangle's actual area.

Recall that the formula for the x of the vertex of a quadratic of the form :

$y=ax^2+bx+c$

is given by the formula:

$x_{vertex}=-\frac{b}{2\,a}$

which in our case gives:

$x_{vertex}=-\frac{b}{2\,a}\\x_{vertex}=-\frac{10}{2\,(-1)}\\x_{vertex}=5$

Therefore, the length of the rectangle is 5, and its width (10-x) is also 5.

The area of the rectangle is therefore the product of these two values: 5 * 5 = 25

Which should coincide with the value we obtain when we replace x by 5 in the area formula:

$y=10x-x^2\\y=50-(5)^2\\y=50-25\\y=25$

8 0

4 years ago

f(x) = 2<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex

loris [4]

Answer:

No answer is possible

Step-by-step explanation:

First, we can identify what the parabola looks like.

A parabola of form ax²+bx+c opens upward if a > 0 and downward if a < 0. The a is what the x² is multiplied by, and in this case, it is positive 2. Therefore, this parabola opens upward.

Next, the vertex of a parabola is equal to -b/(2a). Here, b (what x is multiplied by) is 1 and a =2, so -b/(2a) = -1/4 = -0.25.

This means that the parabola opens upward, and is going down until it reaches the vertex of x=-0.25 and up after that point. Graphing the function confirms this.

Given these, we can then solve for when the endpoints of the interval are reached and go from there.

The first endpoint in -2 ≤ f(x) ≤ 16 is f(x) = 2. Therefore, we can solve for f(x)=-2 by saying

2x²+x-4 = -2

add 2 to both sides to put everything on one side into a quadratic formula

2x²+x-2 = 0

To factor this, we first can identify, in ax²+bx+c, that a=2, b=1, and c=-2. We must find two values that add up to b=1 and multiply to c*a = -2 * 2 = -4. As (2,-2), (4,-1), and (-1,4) are the only integer values that multiply to -4, this will not work. We must apply the quadratic formula, so

x= (-b ± √(b²-4ac))/(2a)

x = (-1 ± √(1-(-4*2*2)))/(2*2)

= (-1 ± √(1+16))/4

= (-1 ± √17) / 4

when f(x) = -2

Next, we can solve for when f(x) = 16

2x²+x-4 = 16

subtract 16 from both sides to make this a quadratic equation

2x²+x-20 = 0

To factor, we must find two values that multiply to -40 and add up to 1. Nothing seems to work here in terms of whole numbers, so we can apply the quadratic formula, so

x = (-1 ± √(1-(-20*2*4)))/(2*2)

= (-1 ± √(1+160))/4

= (-1 ± √161)/4

Our two values of f(x) = -2 are (-1 ± √17) / 4 and our two values of f(x) = 16 are (-1 ± √161)/4 . Our vertex is at x=-0.25, so all values less than that are going down and all values greater than that are going up. We can notice that

(-1 - √17)/4 ≈ -1.3 and (-1-√161)/4 ≈ -3.4 are less than that value, while (-1+√17)/4 ≈ 0.8 and (-1+√161)/4 ≈ 2.9 are greater than that value. This means that when −2 ≤ f(x) ≤ 16 , we have two ranges -- from -3.4 to -1.3 and from 0.8 to 2.9 . Between -1.3 and 0.8, the function goes down then up, with all values less than f(x)=-2. Below -3.4 and above 2.9, all values are greater than f(x) = 16. One thing we can notice is that both ranges have a difference of approximately 2.1 between its high and low x values. The question asks for a value of a where a ≤ x ≤ a+3. As the difference between the high and low values are only 2.1, it would be impossible to have a range of greater than that.

7 0

2 years ago

Football team had scores of 23, 15, 11, 29, and 17 in the last five games the team played. Which score is NOT a prime number?

exis [7]

Answer:

15

Step-by-step explanation:

15 is 5 * 3 so is not prime.

3 0

3 years ago

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