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3 years ago

N < –2 or n ≥ 4 graph

1 answer:

7 0

Make a line graph, plot 0 the go to negative two going to the left open circle and on four going to the roght closed circle

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Please help me asap

klemol [59]

Answer:

1)a

2)d

3)b

4)c

Step-by-step explanation:

8 0

3 years ago

Which number logically follows this series 4 6 9 6 14 6?

Firlakuza [10]

I would say 19.
If you take out the 6s and just look at the rest you have 4, 9, 14.
It is adding 5 each time and just have a 6 between each one.
So logically I would say 19.

3 0

3 years ago

Please help me this is due today

Evgen [1.6K]

Split the room into 2 rectangles
area = 5*4 + 3*3 = 29 sq ft

So the cost is 29*8 = $232

6 0

3 years ago

Find the product of 5(2x) + 3(4y). * Your answer

marysya [2.9K]

Answer:

Step-by-step explanation:

10x + 12y is the correct answer

3 0

2 years ago

Read 2 more answers

Use the Fundamental Theorem of Calculus to find the "area under curve" of

lozanna [386]

Answer:

$\displaystyle A = 300$

General Formulas and Concepts:

Calculus

Integrals

Definite Integrals
Area under the curve
Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

f(x) = 6x + 19

Interval [12, 15]

Step 2: Find Area

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle A = 6(\frac{81}{2}) + 19(3)$
Simplify: $\displaystyle A = 300$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

7 0

2 years ago