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shutvik [7]
3 years ago
10

A number c multiplied by -2 is no less than - 3. Write this word sentence as an inequality.

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer: -3 ≤  -2c

Step-by-step explanation:

First multiply C by -2, and make sure it's not less than -3. It can be -3, but it can't be less than.

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jose jumped 8 1/3 feet. this was 2 2/3 feet farther than lila jumped. how far did lila jump TELL MEH AFTER THIS I GET ON IXL
wariber [46]
Subtract 2 2/3 from 8 1/3

6 1/3
5 0
3 years ago
If y varies inversely as x, and y=1 as x=2 find y for the x- value of -1
Lina20 [59]
Inverse variation is of the form y=k/x  (direct variation is of the form y=kx)

We are given that x=2 when y=1 so using the inverse form y=k/x we can solve for the constant of variation...

1=k/2

k=2

so the equation is:

y=2/x  now they want us to find y when x=-1 so

y=2/-1

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3 0
3 years ago
3. Approximate square root v15
baherus [9]

Answer:

<h2>4</h2>

Step-by-step explanation:

\sqrt{15}\\=3.87298\dots \\\\= 4

4 0
3 years ago
A rectangular room is 3 meters longer than it is wide, and its perimeter is 22 meters. Find the dimension of the room. The lengt
lyudmila [28]

Answer:

Width = 4 m

Length = 7 m

Step-by-step explanation:

given:

perimeter of a rectangle = 22m

L = 3 + W

perimeter = 2L + 2W

perimeter = 2 (3 + W) + 2W

22 = 6 + 2W + 2W

22 - 6 = 4W

W = 16 / 4

W = 4 m

L = 3 + W

L = 3 + 4

L = 7 m

check:

perimeter = 2L + 2W

22 = 2(7) + 2(4)

22 = 14 + 8

22 = 22  ---- OK

4 0
3 years ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
3 years ago
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