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Novosadov [1.4K]
3 years ago
5

Which describes the steps that can be used to solve this problem?

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
3 0
It's b. You have to times $4 by 6 to get $24. Then you take it away from $30 to get the change so $6
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In a geomatric sequence, a5= 36 and a7=16
Anettt [7]
Hello,

a_{5}=36
a_{6}=36*r
a_{7}=36*r^2=16
==>r²=16/36=4/9

==>r=2/3 or r=-2/3


4 0
3 years ago
5 times a number is _ % increase in the number
RSB [31]

Answer:

Answer: Five times a number increased by 7 is 27 from equation Get the answers you need, now!

4 0
3 years ago
Which expressions have a quotient of 4/5​
goblinko [34]

Answer:

  • b and e

Step-by-step explanation:

a. 5/2 ÷ 1/2  =

  • 5/2*2 =
  • 5

b. 6/9 ÷ 5/6

  • 2/3*6/5 =
  • 2*2/5 =
  • 4/5

c. 3/6 ÷ 2/5

  • 1/2 * 5/2 =
  • 5/4

d. 7 1/2 ÷ 6 =

  • 15/2*1/6 =
  • 5/(2*3) =
  • 5/6

e. 1 1/3 ÷ 1 2/3=

  • 4/3 ÷ 5/3 =
  • 4/3 * 3/5 =
  • 4/5
5 0
2 years ago
Read 2 more answers
Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
Read 2 more answers
Twenty people each had 36 carrots. How many carrots are there altogether?
elena-s [515]

You know that there are 25 people and each person had 36 carrots. To find how many carrots everyone has in total, you have to multiply 25 by 36.

25 times 36 is 900.

There are 900 carrots altogether.

7 0
2 years ago
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