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3 years ago

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The hypotenuse of a right triangle is three times the length of its first leg. The length of the other leg is four feet. Find th

e lengths of the first leg and the hypotenuse and enter them in the below spaces in this order. For non-integer answer(s), round your answer(s) to the nearest tenth.

1 answer:

ioda3 years ago

5 0

Answer:

The length of hypotenuse is 4.2 feet and length of first leg is 1.4 feet.

Step-by-step explanation:

Consider the provided information.

Let x is the length of the first leg.

It is given that the hypotenuse of a right triangle is three times the length of its first leg.

Hypotenuse = 3x

The length of the other leg is 4 feet.

Pythagorean theorem.

$a^2+b^2=c^2$

Where a and b are the legs and c is the hypotenuse.

Substitute the respective values in the above formula.

$x^2+4^2=(3x)^2$

$x^2+16=9x^2$

$16=8x^2$

$2=x^2$

$\sqrt{2}=x$

x = 1.4 feet(nearest tenth)

The length of hypotenuse = 3x = 3×1.4 = 4.2 feet

The length of hypotenuse is 4.2 feet and length of first leg is 1.4 feet.

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Which statements about right triangles and hexagons are true?

Goryan [66]

Hello :) = Merhaba :)

Answer:

<h3><u><em>A and C</em></u></h3>

Step-by-step explanation:

Let's look at the chances:

A) Both shapes are polygons.

This answer is true. Because the triangle and the hexagon are polygons.

B) The first shape has the same number of sides as the second shape.

This answer is false. Because three sides of the triangle have six edges of the hexagon.

C) Both shapes have all sides of equal length.

This answer is true. Because the smooth polygons have equal edges.

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3 years ago

Sam and Alex play Tennis. On the weekend Sam played 8 more games than Alex did, and together they played 24 games. How many game

Tresset [83]

It's 4 ..........................

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3 years ago

−9 ≤ 3m − 4 < 11, m ∈ I

zloy xaker [14]

Answer:

$\frac{-5}{3} \leq m < 5$

Step-by-step explanation:

Isolate the 3m by adding 4 to every part of the inequality:

-9 ≤ 3m - 4 < 11

-9 + 4 ≤ 3m -4 + 4 < 11 + 4

-5 ≤ 3m < 15

Now, divide every part of the inequality by 3:

$\frac{-5}{3} \leq \frac{3m}{3} < \frac{15}{3}$

$\frac{-5}{3} \leq m < 5$

Hope this helps!

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3 years ago

A California grower has an 80-acre farm on which to plant strawberries and tomatoes. The grower has available 600 hours of labor

Makovka662 [10]

Answer:

a)

P=500s+300t

$s+t \leq 80$

$s\leq 60$

$T \leq 50$

$8s+4t \leq 600$

$5s+20t \leq 800$

$s \geq 0$

$t \geq 0$

b) See attached picture.

Possible extreme points:

(0,0), (0,40), (160/3, 80/3), (60,20), (60,0)

c) Optimal solution points:

(0,0), (0,40), (160/3, 80/3), (60,20), (60,0)

Optimal objective function value happens at (60,20) for a profit of $36,000.

Step-by-step Explanation:

a) In order to find the linear programming model, we need to start by setting our variables up.

s= number of acres of strawberries to plant.

t = number of acres of tomatos to plant.

With this we can start by setting our ojective function up, which is the profit function.

"The profit from an acre of strawberries is $500, and the profit from an acre of tomatoes is $300."

Our objective function is:

P=500s+300t

Next, we need to set our restrictions up, which come from the rest of the sentences of the problem:

"A California grower has an 80-acre farm on which to plant strawberries and tomatoes." This tells us that we can plant as much as 80 acres of strawberries and tomatoes, so our first restriction is:

$s+t \leq 80$

next:

"...and he has contracted for shipping space for a maximum of 60 acres’ worth of strawberries and 50 acres’ worth of tomatoes."

This sentence gives us the next two restrictions:

$s \leq 60$

$T \leq 50$

"The grower has available 600 hours of labor per week...", " An acre of strawberries requires 8 hours of labor..." and "...an acre of tomatoes requires 4 hours of labor...".

These sentences give us the next restriction:

$8s+4t \leq 600$

"The grower has available...800 tons of fertilizer...", "An acre of strawberries requires...5 tons of fertilizer", "...an acre of tomatoes requires...20 tons of fertilizer."

with this information we can build our next restriction:

$5s+20t \leq 800$

and finally, we also know that we cannot plant less than 0 acres of tomatos or strawberries since that would become a loss. So the final restrictions are:

$s \geq 0$

$t \geq 0$

So the linear programmin model is the following:

P=500s+300t

$s+t \leq 80$

$s \leq 60$

$T \leq 50$

$8s+4t \leq 600$

$5s+20t \leq 800$

$s \geq 0$

$t \geq 0$

b) Once we have the linear programming model, we can go ahead and graph each of the restrictions. All the restrictions are graphed the same so I will give a brief explanation on how to graph the first one.

So let's take the first restriction:

$s+t \leq 80$

we can start by turning it into an equation:

s+t=80

and pick any value we wish for s. We can pick s=0 since that will simplify the work:

0+t=80

therefore t=80

the first point to plot is (0,80)

in order to find the second point to plot we can set t=0 so we get:

s+0=80

s=80

therefore the second point to plot is (80,0)

We can plot these two points on our coordinate axis and connect them with a solid line.

Next, we know that the region to shade should be less than or equal to 80, so we pick a test point above and below the graph. Let's pick (100,60) and (0,0)

for (100,60) we get that:

$100+60 \leq 80$

$160 \leq 80$

this is false so that region should not be shaded. Let's take the other test point:

for (0,0) we get:

$0+0\leq 80$

$0 \leq 80$

is true, so we should shade the region below the graph.

The same procedure is done with the rest of the restrictions and we have as a result the graph in the attached picture.

The feasible area is the area all the shaded areas have in common.

The extreme points are the vertices of the polygon formed by the feasible area, we can find them graphically or algebraically.

If we were to find them algebraically we would solve the corresponding system of equations. The first point (0,0) is found at the intersection of the restrictions: $s \geq 0$ and $t \geq 0$ .

The second extreme point is at the intersection of the restrictions: $s \geq 0$ and $5s+20t \leq 800$ , which yields (0,40).

The next extreme point is at the intersection of the restrictions: $5s+20t \leq 800$ and $s+t \leq 80$ . When solving this system of equations we get the point: (160/3, 80/3).

The next extreme point is at the intersection of the restrictions: $s+t \leq 80$ and $s \leq 60$ which yields (60,20)

and the final extreme point is at the intersection of the restriction: $s \leq 60$ and $t \geq 0$ which yields (60,0)

so the possible extreme points are: (0,0), (0,40), (160/3, 80/3), (60,20), (60,0).

c) Now we solve the model, in order to solve the model we need to use the optimal solution points and evaluate them in the objective function:

P=500s+300t.

See attached table for the results of substituting the optimal points.

So the optimal point will happen at (60,20) with a profit of $36,000.

5 0

3 years ago

Suppose you want to transform the graph of the function into the graph of the function . Which transformations should you perfor

nexus9112 [7]

Answer:

A. Reflect the graph of the first function across x-axis, translate it $\frac{\pi}{4}$ to the left, and translate it 2 units up.

Step-by-step explanation:

We have the original function is $y=\tan (x+\frac{\pi}{4})-1$ .

The new transformed function is given by $y=-\tan (x+\frac{\pi}{2})+1$ .

So, we can see that the following sequence of transformations have been applied to the original function:

1. The function f(x) is reflected about x-axis i.e. f(x) becomes -f(x), which gives $y=-\tan (x+\frac{\pi}{4})-1$

2. This function obtained is translated $\frac{\pi}{4}$ units to the left i.e. $y=-\tan (x+\frac{\pi}{4}+\frac{\pi}{4})-1$ i.e. $y=-\tan (x+\frac{\pi}{2})-1$

3. Finally, this new function is translated 2 units upwards i.e. $y=-\tan (x+\frac{\pi}{2})-1+2$ i.e. $y=-\tan (x+\frac{\pi}{2})+1$

Hence, after applying, 'reflection across x-axis, translation of $\frac{\pi}{4}$ to the left, and translation of 2 units up', we get the required function.

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