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3 years ago

Use compatible numbers to estimate the quotient394.6 divided by 9

Mathematics

1 answer:

laiz [17]3 years ago

8 0

Answer:

Step-by-step explanation:

As a first rough estimate: divide 390 by 10, obtaining 39.

Another reasonable approximation could be obtained by dividing 450 by 9 and 360 by 9, obtaining 50 and 40. Since 390 is roughly halfway between 360 and 450, estimate that the quotient is approx 45 (halfway between 40 and 50).

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I WILL GIVE BRILLIANT

Katyanochek1 [597]

Answer:

The answer is 43.27, 27, and 10.58

Step-by-step explanation:

First you add 24.39 to 5.99 and get 30.38 then add 12.89 to that and get 43.27.

The second problem you add 15.4 to 11.6 and get 27 degrees.

The third one you subtract 16.92 from 27.5 and get 10.58.

7 0

3 years ago

HƏĻP MƏ Evaluate the expression when f= 125, 3/5(F - 20)

Lilit [14]

Answer:

Step-by-step explanation:

plug 125 in for 'f'

3/5(125-20) = 3/5 · 105 = 315/5 = 63

5 0

3 years ago

X divided by 7 - 5 = 2 x/7 - 5 = 2

Black_prince [1.1K]

Answer:

x=4

Step-by-step explanation:

Multiply by 2 from both sides of equation.

2x/7-5=2*2

Simplify, to find the answer.

2*2=4

x=4 is the correct answer.

I hope this helps you, and have a wonderful day!

4 0

4 years ago

Which of the following equations has the same solutions as x^2+8x+2=0

denis23 [38]

Answer:

See below

Step-by-step explanation:

Need the choices.....

essentially, you can multiply this equation by any constant to get an equivalent equation with the same solutions.

x^2 + 8x + 2 = 0 multiply both sides by '3' to get

3x^2 + 24 x + 6 = 0 <===== this will have the same solutions

or this one:

x^2 + 8x + 2 = 0 Multiply both sides by '6' to get

6x^2 + 48x + 12 = 0 <===== this will have the same solutions also

etc....

7 0

2 years ago

The velocity function, in feet per second, is given for a particle moving along a straight line. Find (a) the displacement and (

AveGali [126]

Answer:

(a) 2 feet.

(b) 2 feet.

Step-by-step explanation:

We have been given that the velocity function $v(t)=\frac{1}{\sqrt{t}}$ in feet per second, is given for a particle moving along a straight line.

(a) We are asked to find the displacement over the interval $1\leq t\leq 4$ .

Since velocity is derivative of position function , so to find the displacement (position shift) from the velocity function, we need to integrate the velocity function.

$\int\limits^b_a {v(t)} \, dt$

$\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$

$\int\limits^4_1 {\frac{1}{t^{\frac{1}{2}}} \, dt$

$\int\limits^4_1 t^{-\frac{1}{2}} \, dt$

Using power rule, we will get:

$\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}}\right] ^4_1$

$\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}}\right] ^4_1$

$\left[2t^{\frac{1}{2}}\right] ^4_1$

$2(4)^{\frac{1}{2}}-2(1)^{\frac{1}{2}}=2(2)-2=4-2=2$

Therefore, the total displacement on the interval $1\leq t\leq 4$ would be 2 feet.

(b). For distance we need to integrate the absolute value of the velocity function.

$\int\limits^b_a |{v(t)|} \, dt$

$\int\limits^4_1 |{\frac{1}{\sqrt{t}}}| \, dt$

Since square root is not defined for negative numbers, so our integral would be $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ .

We already figured out that the value of $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ is 2 feet, therefore, the total distance over the interval $1\leq t\leq 4$ would be 2 feet.

7 0

3 years ago