All categories

3 years ago

HELP PLEASEEEEEEEEEeeeee

Mathematics

1 answer:

maks197457 [2]3 years ago

3 0

Might have to experiment a bit to choose the right answer.

In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.

Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456

In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.

Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.

You might be interested in

A cookie factory uses 1/2 of a barrel of oatmeal in each batch of cookies. The factory used 3 1/2 barrels of oatmeal yesterday.

Helen [10]

First, to make this a bit easier, we can turn the fractions into decimals.

1/2 = 0.5 , 3 1/2 = 3.5

Divide 3.5 by 0.5.

3.5 ÷ 0.5 = 7

<h2>Answer: </h2>

The factory made 7 batches of cookies.

7 0

3 years ago

What is the equation of the line that is paralle to y=3x-8 and passes thur the point (4,-5)

Brrunno [24]

⊰_________________________________________________________⊱

Answer:

The equation is-: y=3x

Step-by-step explanation:

$\large\displaystyle\text{$\begin{gathered} \sf{Substitute \ the \ values \ into \ the \ formula \ y-y_1=m(x-x_1)} \\ \sf {parallel \ lines \ have \ same \ slopes, \ thus} \\ \sf{slope \ of \ the \ 2nd \ line = 3}\\ \sf{now \ substitute \ the \ values} \\ \sf {y-(-5)=3(x-4)}\\ \sf{y+5=3(x-4) (It's \ Point-Slope\;Form, \ see \ below \ for \ slope-intercept)}\\ \sf {y+5=3x-12} \\ \sf{y=3x-12-5} \\ \sf{y-3x-17} \end{gathered}$}}$

$\pmb{\tt{done \ !!}}$

⊱_________________________________________________________⊰

4 0

2 years ago

Triangle E D F has sides that are of varying lengths. Side E D has a length of 4 and side E F has a length of 9.

9966 [12]

Answer:

D E + E F greater-than D F

5 less-than D F less-than 13

Triangle D E F is a scalene triangle

Step-by-step explanation:

we know that

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side

we have the triangle EDF

where

$ED=4\ units\\EF=9\ units$

Applying the triangle inequality theorem

$ED+EF > DF\\4+9 > DF\\13 > DF\\DF < 13\ units$

$ED+DF > EF\\4+DF > 9\\DF > 5\ units$

The length of DF is the interval -----> (5,13)

The triangle DEF is a scalene triangle (the three length sides are different)

therefore

The statements that are true are

D E + E F greater-than D F

5 less-than D F less-than 13

Triangle D E F is a scalene triangle

3 0

3 years ago

Read 2 more answers

The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.

puteri [66]

Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Area of a Rectangle: A = lw

l is length
w is width

Calculus

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

$\displaystyle \frac{dl}{dt} = 4 \ m/day$

$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

[Area of Rectangle] Product Rule: $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

[Rate] Substitute in variables [Derivative]: $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
[Rate] Multiply: $\displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day$
[Rate] Add: $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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

Unit: Implicit Differentiation

Book: College Calculus 10e

8 0

3 years ago

Solve the system of equations by substitution

fenix001 [56]

I hope you can read my writing and hope this helps

4 0

2 years ago