Ask question

Ask question

All categories

3 years ago

13

Is this a liner equation ,the price of a loaf of bread increases by .25 each week

1 answer:

alexgriva [62]3 years ago

5 0

i think its not because it should increase so it would go diagonal

You might be interested in

What is the factorization of 121b4 − 49? (11b − 7)(11b − 7) (11b 7)(11b − 7) (11b2 − 7)(11b2 − 7) (11b2 7)(11b2 − 7)

Inessa05 [86]

The Factorization of 121b⁴ − 49 is (11b^2 + 7)(11b^2 - 7).

The equation 121b⁴ − 49

To find the Factorization of 121b⁴ − 49.

<h3>What is the factor of a^2-b^2?</h3>

The factor of a^2-b^2 is (a+b)(a-b)

We have write the given equation in the form of a^2-b^2

$= 121b^4 - 49\\= (11b^2)^2 - 7^2\\= (11b^2 + 7)(11b^2 - 7)$

Therefore the factor of the 121b^4 − 49 is (11b^2 + 7)(11b^2 - 7).

To learn more about the factor visit:

brainly.com/question/25829061

5 0

2 years ago

5.A Nathan decided he wanted to go to a theme park. The theme

rjkz [21]

Answer:

5a. $1.50R + $7

5b. 14 rides

Step-by-step explanation:

5b. $28 - $7= $21/1.50= 14

He can go on 14 rides

3 0

2 years ago

Kelsey had $65 to spend on books. Each book cost $5.50, and there was a $7.50 fee for shipping. She let b equal the number of bo

larisa86 [58]

Kelsey must spend less than or equal to $65.
The expression 5.50b+7.5 is correct
Kelsey able to purchase 10 books.
On the end Kelsey can spend $2.50 to purchase small present (calendar or book for notes)

6 0

3 years ago

Read 2 more answers

Factor each polynomial completely. If you use a Special Pattern, identify which one you used.

VARVARA [1.3K]

Answer:

6,true 7,false 8,false 9,true 10,true

Step-by-step explanation:

8 0

3 years ago

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

5 0

3 years ago