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Salsk061 [2.6K]

2 years ago

7

A gardener is installing fence around his garden. Let x represent the width of the garden, in feet. The perimeter of the garden

is 8x + 8. Which expression represents the length of the garden?

1 answer:

ivanzaharov [21]2 years ago

3 0

Answer:

(8) x+8 in (8)

Step-by-step explanation:

width times length

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The coefficients corresponding to k = 0, 1, 2, ..., 6 in the expansion of (x + y)^6 are _____.

Kryger [21]

Using the triangle of pascal we have that the expression equivalent to (x + y) ^ 6 is given by:
x ^ 6 + 6x ^ 5y + 15x ^ 4y ^ 2 + 20x ^ 3y ^ 3 + 15x ^ 2y ^ 4 + 6xy ^ 5 + y ^ 6
Therefore, the coefficients of the expansion are given by:
1, 6, 15, 20, 15, 6, 1
Answer:
The coefficients corresponding to k = 0, 1, 2, ..., 6 in the expansion of (x + y) ^ 6 are 1, 6, 15, 20, 15, 6, 1

5 0

3 years ago

Read 2 more answers

Solve for X. Please!!

pychu [463]

Answer:

x = 37.5

Step-by-step explanation:

By Basic Proportionality Theorem:

$\frac{x}{20} = \frac{46 - 16}{16} \\ \\ \frac{x}{20} = \frac{30}{16} \\ \\ x = \frac{20 \times 30}{16} \\ \\ x = \frac{600}{16} \\ \\ x = 37.5$

3 0

2 years ago

The admissions office of a private university released the following admission data for the preceding academic year: From a pool

lina2011 [118]

Answer: Our required probability is 0.1695.

Step-by-step explanation:

Since we have given that

Number of male applicants = 4200

Number of female applicants = 3800

So, total number of applicants = 4200+3800 = 8000

Probability of male entered and subsequently enrolled is given by

$0.4\times 0.4=0.16$

Probability of female entered and subsequently enrolled is given by

$0.4\times 0.45\\\\=0.18$

Number of male entered and subsequently enrolled is given by

$0.16\times 4200\\\\=672$

Number of female entered and subsequently enrolled is given by

$0.18\times 3800\\\\=684$

So, Probability that a student who applied for admission will be accepted by the university and subsequently will enroll is given by

$\dfrac{672+684}{8000}\\\\=\dfrac{1356}{8000}\\\\=0.1695$

Hence, our required probability is 0.1695.

3 0

2 years ago

The length of a rectangular field is 7 m less than 4 times the width. The perimeter is 136 m. Find the width and the length

Helga [31]

Answer:

Step-by-step explanation:

Represent the width by W. Then, "The length of a rectangular field is 7 m less than 4 times the width" expressed symbolically is

L = 4W - 7 (dimensions in meters)

Recall that the perimeter formula in this case is P = 2L + 2W, and recognize that the perimeter value is 136 m. After substituting 4W - 7 for L, we get:

136 m = 2(4W - 7) + 2W, or

136 = 8W - 14 + 2W, or

150 = 10W These three equations are equivalent mathematical statements.

150 = 10W reduces to W = 15 (meters).

Part A: the independent variable is W, the width of the field.

Part B: The mathematical statement is 136 m = 2(4W - 7) + 2W, which after algebraic manipulation becomes 150 = 10W.

Part C: The above equation can be solved for W: W = 15 meters. This is the value of the independent variable.

6 0

2 years ago

A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order

Mazyrski [523]

Answer:

Feed 1, Soybean meal required = $\dfrac{2000}{3} \text{ lbs.}$

Feed 2, Corn meal required = $\dfrac{4000}{3} \text{ lbs.}$

Step-by-step explanation:

Total feed is <em>1 ton</em> i.e. <em>2000 lbs</em>.

Let <em>x</em> be the amount of Feed 1 required.

Feed 1 has $45\%$ of protein.

$\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)$

Then, amount of Feed 2 required = $(2000 - x) \text{ lbs.}$

Feed 2 has $15\%$ of protein.

$\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)$

As per question, total protein required is $25\%$ of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

$\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}$

Feed 1 required = $\dfrac{2000}{3}\text{ lbs .}$

Feed 2 required = $2000 - \dfrac{2000}{3}\\$

$\Rightarrow \dfrac{4000}{3}\text{ lbs.}$

7 0

3 years ago