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3 years ago

I just need the straight up answer lolol

Mathematics

1 answer:

Licemer1 [7]3 years ago

8 0

The correct answer is -5

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Determine if the sequence is arithmetic. If it is, find the common difference. Is the sequence a function ?

BaLLatris [955]

$6)\\r=a_{n+1}-a_n\to r=a_2-a_1=a_3-a_2=a_4-a_3=...\\\\a_1=-7;\ a_2=-10;\ a_3=-11;\ a_4=-13\\\\a_2-a_1=-10-(-7)=-10+7=-3\\a_3-a_2=-11-(-10)=-11+10=-1\\\\a_3-a_2\neq a_2-a_1\\\\\text{It's not an arithmetic sequence}$

$\r=\dfrac{a_{n+1}}{a_n}\\\\r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...\\\\8.)\\a_1=4;\ a_2=20;\ a_3=100;\ a_4=500\\\\r=\dfrac{20}{4}=\dfrac{100}{20}=\dfrac{500}{100}=5\\\\10.)\\a_1-30;\ a_2=-45;\ a_3=-50;\ a_4=-65\\\\\dfrac{a_2}{a_1}=\dfrac{-45}{-30}=\dfrac{3}{2}\\\\\dfrac{a_3}{a_2}=\dfrac{-50}{-45}=\dfrac{10}{9}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}$
$12.)\\a_1=-7;\ a_2=-14;\ a_3=-21;\ a_4=-28\\\\\dfrac{a_2}{a_1}=\dfrac{-14}{-7}=2\\\\\dfrac{a_3}{a_2}=\dfrac{-21}{-14}=\dfrac{3}{2}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}\\\\\text{It's a function}$

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2 years ago

Here are the ingrediants needed make 8 pancakes.

vazorg [7]

I think Craig would need 7.5g of butter

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3 years ago

Amelia drives 10 miles in 20 minutes. If she drove three hours in total at the same rate, how far did she go? subquestion - how

olga nikolaevna [1]

Answer:

Amelia drove 90 miles in 3 hours.

Step-by-step explanation:

she drove 10 miles in 20 minutes

we multiply this by 3, and get

30 miles in an hour ( 60 minutes)

then we multiply this by 3 and get

90 miles in 4 hours (180 minutes) :)

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3 years ago

Hello I need help<br> Find the missing side lengths, leave your answers as radicals in simplest form

Jet001 [13]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$y = \frac{1}{2} \times \frac{8 \sqrt{3} }{3} \\$

$y = \frac{4 \sqrt{3} }{3} \\$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$x = \frac{ \sqrt{3} }{2} \times \frac{8 \sqrt{3} }{3} \\$

$x = \frac{8 \times { \sqrt{3} }^{2} }{2 \times 3} \\$

$x = \frac{8 \times 3}{2 \times 3} \\$

$x = \frac{8}{2} \\$

$x = 4$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

4 0

2 years ago

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

2 years ago