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natali 33 [55]
3 years ago
7

Which expressions are equivalent to 2(4f + 2g) ?

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
5 0

Answer:

8f+4g

Step-by-step explanation:

distribute the 2 to both 4 and 2 in the the bracket

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The bull’s-eye on a target has diameter 2 in. What is the area of the bull’s-eye to the nearest tenth? Use 3.14 for π
777dan777 [17]

Answer:

20.3cm^2

Step-by-step explanation:

Bullseye has a circular shape. So area of a bulleyse = Pi x (radius) ^2

Díameter is 2 inches = 5. 08cm

But radius = diameter /2

Radius = 5.08cm /2

Radius = 2.54cm

Hence Area of bulleyse = 3.14 x (2.54) ^2

= 3.14 x 6. 451 cm^2

= 20.256 cm^2

=20.3cm^2 (to nearest tenth)

3 0
3 years ago
How to evaluate the expression
noname [10]
[(-72) \div (-2)]^3 - (-7) \times (-10)

= [36]^3 - (-7) \times (-10)

=46656 - (-7) \times (-10)

=46656 - (70)

=46586

-------------------------------------


2^3 - 9 \times (2 - 4)

= 8 - 9 \times (-2)

= 8 - (-18)

= 8 + 18

= 26

4 0
3 years ago
Use rounding or compatible numbers to estimate the sum 271 + 425
zysi [14]

Answer:

700

Step-by-step explanation:

For this one I'll do estimates. 271 estimated would drop down to 270. And 425 would bump up to 430. Adding 270 and 400 you would get a round number of 700!

Hope this helped.

4 0
3 years ago
Read 2 more answers
The universal ga law PV=net describes the relationship among the pressure volume and temperature of a gas.
miskamm [114]

Answer:

R ≈ 8.31

Step-by-step explanation:

Filling in the given values ...

... pV = nRT

... 2·4.155 = 1·R·1

... 8.31 = R

7 0
3 years ago
A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is
Monica [59]

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

l^2 = h^2 +d^2

l^2 = 6^2 + d^2

l^2 = 36 + d^2

We relate to:  2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}

From the question;

l = 34,

So to find \dfrac{dd}{dt}, we get;

d = \sqrt{l^2 - 36}

d = \sqrt{34^2 - 36}

d = \sqrt{1156- 36}

d = \sqrt{1120}

d = 33.46

So, we have:

\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}

\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5

\dfrac{dd}{dt}= 1.016 \times - 5

\dfrac{dd}{dt}=-5.08 \ ft/sec

4 0
3 years ago
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