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2 years ago

All points of the step function f(x) are graphed.

Mathematics

1 answer:

slamgirl [31]2 years ago

5 0

Answer:

(a) {x| –4 < x ≤ 4}

Step-by-step explanation:

The domain is the horizontal extent of the graph.

Your problem statement tells you the left end is an open circle at x=-4, and the right end is a closed circle at x=4. The domain is ...

-4 < x ≤ 4

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Identify next three numbers in this sequence: 2, 6, 3, 9, 6, 18, 15

kherson [118]

Answer:

45,42,126

multiply by 3

6 0

3 years ago

Marco buys a certain brand of shampoo for a supplier $7.25 per bottle he sells it to his customers at a makeup 25% what would th

SCORPION-xisa [38]

Answer:

Markup= 1.81

Sale price =9.06

Step-by-step explanation:

The markup is the original prices times the markup rate

markup = 7.25 * 25%

= 7.25 * .25

=1.8125

We round this to the nearest cent

=1.81

The new sale price will be the original price + the markup

7.25+1.81

9.06

3 0

3 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago

Kain graphs the hyperbola (y+2)^2/64 − (x+5)^2/36 = 1 . How does he proceed? Drag a value, phrase, equation, or coordinates in t

cupoosta [38]

Answer:

1st box: (-5,-2)

2nd box: up and down

3rd box: 8

4th box: +-4/3

5th box: y+2 = +-4/3 (x+ 5)

Step-by-step explanation:

Just took the k12 unit test, hope this helps!

7 0

2 years ago

Read 2 more answers

The yield of orange trees is reduced if they are planted too close together. If there are 30 trees per acre, each tree produces

Tanya [424]

Answer:

5 more than 30 trees should be planted, for a total of 40 trees per acre.

Step-by-step explanation:

Let x be the number of trees beyond 30 that are planted on the acre

The number of oranges produced = Oranges(x) = (number of trees) (yield per tree)

We are given that For each additional tree in the acre, the yield is reduced by 7 oranges per tree

So, number of oranges produced = $(30 + x)(400 -10x)$

= $12000 -300x+400x-10x^2$

= $12000 + 100x-10x^2$

The derivative Oranges'(x) = 100-20x

Substitute first derivative equals to 0

$100-20x =0$

$x=5$

Using the second derivative test,

Oranges"(x) = -20

20 is negative,

So, this is the case of maximum.

Thus, 5 more than 30 trees should be planted, for a total of 40 trees per acre.

4 0

3 years ago