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Vaselesa [24]
2 years ago
15

How many inches are in 12 1/2 feet

Mathematics
2 answers:
shusha [124]2 years ago
7 0

Answer:

150 inches

Step-by-step explanation:

12.5 x 12 = 150 inches

Delvig [45]2 years ago
5 0

Answer:

The answer is 150 inches

Step-by-step explanation:

Multiply the length value by 12

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2900 dollars is placed in an account with an annual interest rate of 9% Wirte an equation that calculates how many dollars will
valentinak56 [21]

Answer:

The amount that will be in the account will be $(2900 +261x)

Step-by-step explanation:

We know that the amount of money that will be in the account will be a sum of the initial principal ($2900) and the interest that is accrued after x years.

Since we already know the principal, we will have to calculate the interest that is accrued after x years.

we can use the formula:

Interest = (principal X rate X time)/ 100

Interest = (2900 \times 9 \times x)/100=261x

Therefore the amount that will be in the account will be $(2900 +261x)

6 0
3 years ago
What two fractions can I multiply to make 2/3?
chubhunter [2.5K]

Answer:

2/3*3/3

Step-by-step explanation:

I mean its just doing 2/3 * 1 which i mean, 3/3 is still a fraction so i did it (TBH cant think of anything else

3 0
2 years ago
Read 2 more answers
(2x + 35)=(5x-22) what’s x
finlep [7]

Answer:

x=19

Step-by-step explanation:

5 0
3 years ago
What fraction of a day is 15 hours
Genrish500 [490]
15/24 = 5/8


15 hours out of 24 hours.!!!!


5/8 of a day, or 0.625 as a decimal.


in minutes, it's 900/1,440



Hope this helps!!!!!
5 0
3 years ago
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
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