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2 years ago

What are these rounded to the nearest tenth help 1)1017.36 2)1139.82

Mathematics

2 answers:

ziro4ka [17]2 years ago

8 0

Answer:

1. 1017.4

2. 1139.8

Step-by-step explanation:

Those the answers

Mazyrski [523]2 years ago

7 0

Answer: 1,017.4 and 1,139.8

Step-by-step explanation:

If the digit after tenth is greater than or equal to 5, add 1 to tenth. Else remove the digit.

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Ad libitum [116K]

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Step-by-step explanation:

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2 years ago

Read 2 more answers

77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

Colt1911 [192]

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$ .

6 0

1 year ago

If 5y + 3 = 2y + 9 then find 2y=

Serhud [2]

If you were to solve for y, it would be 2. Therefore 2y would be 4.

5 0

3 years ago

A random sample of 11 days were selected from last year's records maintained by the maternity ward in a local hospital, and the

EleoNora [17]

Answer: ( Min = 0 , $Q_1=1$ , $Q_2=3$ , $Q_3=7$ , Max = 10 )

Step-by-step explanation:

Given : A random sample of 11 days were selected from last year's records maintained by the maternity ward in a local hospital, and the number of babies born each day of the days is given below:

3 7 7 10 0 7 1 2 5 3 0

We first arrange them in increasing order , we get

0 0 1 2 3 3 5 7 7 7 10

Here , N= 11

Now , we can see that

Minimum value = 0

Maximum value = 10

First quartile $Q_1$ = $(\dfrac{N+1}{4})^{th}\ term=(\dfrac{12}{4})^{th}\ term = 3^{rd} term =1$