Evaluate the expression y2 when y=2.5 .

A rectangular soccer field is twice as long as it is wide. If the perimeter of a soccer field is 300 yards, what are the fields'

Masteriza [31]

A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards , what are its dimensions?
I know the basic formula is 2W+2L=300 but i am not sure where to go from there...
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Equations:
2W + 2L = 300
L = 2W
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Substitute for "L" and solve for "W":
2W + 2(2W) = 300
6W = 300
W = 50 yds (width)
----
Solve for "L":
L = 2W
L = 100 yds (length)
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Cheers.

8 0

2 years ago

I need help lol!!

zmey [24]

Answer: Its 7,6 HONEYZ

Step-by-step explanation:

4 0

2 years ago

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

On integration , we get

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$

After simplification and cancellation we get

= $\frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6$

on calculation , we get

= $\frac{124,416}{5} + \frac{864}{3} + 180$

On L.C.M 15

= $\frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}$

= 25 351.2 units²

Conclusion:-

The area of the region between the graph of the given function and the x-axis = 25,351 units²

6 0

2 years ago

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

or

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .

Observe the image below.

7 0

3 years ago

Im not sure if anyone can help me but if you can i would really appreciate it cuz im really confused about this so thank you. al

Advocard [28]

Answer:

Question 1) 6x+5

Question 2) Blank #1: sum. Blank #2: 5. Blank #3: 4

See work in the attachment. If you're still confused, read the step-by-step explanation for a more in-depth explanation of what exactly I did for my work. Let me know if you have any questions :>

Step-by-step explanation:

For question 1, it's important to know that the perimeter is equal to the sum of all the sides. Since the shape is a rectangle, the sides opposite of each other are the same length. Thus, the left and right sides are both 3x - 3. 2(3x - 3) = 6x - 6. The unknown value, which I called a, is the same for both the top and the bottom. To find it, you must first subtract 6x - 6 from the perimeter of 18x + 4. This equals 12x + 10. Then, divide that answer by 2 to get the unknown side. You then get 6x + 5 as the answer for the unknown side.

To make sure it's correct, you can plug in a random number for x. In this case, I checked my work by plugging 2 in for x.

Perimeter = 18(2) + 4 = 36 + 4 = 40

Perimeter = 6(2) + 5 + 6(2) + 5 + 3(2) - 3 + 3(2) - 3 = 12 + 5 + 12 + 5 + 6 - 3 + 6 - 3 = 24 + 10 + 12 - 6 = 40

For question 2, the main idea you need to know is that A = lw. The area of the first rectangle is 6 * 5 = 30. The area of the second rectangle is 6 * 4 = 24. When you add them together, you get 54. The area of the rectangle that is formed when you combine rectangle 1 and rectangle 2 is 6 * 9 = 54. This shows that 6(4) + 6(5) = 6(9)

6 0

2 years ago