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Elena-2011 [213]

3 years ago

6

if the perimeter of Milo's rectangular backyard Is 16 feet. which of the following could be the dimensions of the yard? circle a

ll that apply. explain your choice

1 answer:

postnew [5]3 years ago

6 0

Answer:

the answer is a and d

Step-by-step explanation:

6 + 6 + 2 +2 = 16

3 + 3 + 5 + 5 = 16

to find perimeter, double each factor and add :)

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Please solve these. Will rate you 5 star!

ziro4ka [17]

Answer: Hi!

To find the answer for question 24, we would first find the square root of 20 and 5. The square root of 20 is about 4.47 (I used a calculator for convenience) and the square root of 5 is about 2.24. Now we need to add these:

4.47 + 2.24 = 6.71

Now we divide 30 by this:

30/6.71 = 4.47

Now, we'll evaluate the answer choices.

a) is equal to 1.49

b) is equal to 13.4

c) is equal to 4.47

d) is equal to 26.88

The answer choice that is correct is c, because it is what we got for the problem that we were supposed to solve in the beginning.

Hope this helps!

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3 years ago

mrs wong manages a restaurant which is currently hiring. if today she interviewed 4 waiters and 2 chefs, how like the first appl

Yanka [14]

4/6 chance or 2/3 chance. i’m not sure if this is like probability i’m at a middle school level

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3 years ago

3/5 of a Conductors Club went on the last camping trip.If 39 people went on the trip, how many are in the club?

Lerok [7]

The answer is 65 people in the club.

4 0

4 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]

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3 years ago

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brilliants [131]

Answer:

111

Step-by-step explanation:

il help u if u help me look at my thing and ill help u

7 0

3 years ago