Vashti and Stacy measure the width of the gym at their school. It is 17 yards wide.

Which of these number words represents the numeral 4,056? four thousand, five hundred six four thousand, five hundred fifty-six

hjlf

Solution:

we have been asked to find

Which of the given number words represents the numeral 4,056?

Its about the counting.

On unit place we have 6.

on Tenth place we have 5.

on hundredth place we have 0.

On Thousand place we have 4.

so in our number word representation there will be no "Hundreds".

So it will make

Hence the given number words representing the numeral 4,056 is Four thousands and fifty six

4 0

3 years ago

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What is the measure for the following ∠? m∠BOX 20 40 60

Daniel [21]

Answer:

40

Step-by-step explanation:

3 0

2 years ago

Hello I need help with question B and C

Tanzania [10]

Answer: a) 22.94 psi

b) 5.93 X 10 --5

Step-by-step explanation:

a)The pressure at which will trigger a warning is

31 - 31*0.26 = 22.94 psi

b) The probability that that the TPMS will trigger warning at 22.94 psi, given that tire pressure has a normal distribution with average of 31 psi and standard deviation of 2 psi

where x = 22.94,

do not nothe rest

6 0

2 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

2 years ago

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How many ways are there to choose a half dozen donuts from 10 varieties a)If there are no two donuts of the same variety.b)If th

777dan777 [17]

Answer:

If multiples are allowed (like "10 plain, two chocolate"), then for each of 12 donuts there are 21 possibilities.

That is 21 x 21 x 21 .... x 21 (12 terms in all) = 21^12

Step-by-step explanation:

7 0

3 years ago