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Hitman42 [59]
3 years ago
11

can someone please help me with this math problem?

Mathematics
1 answer:
Ugo [173]3 years ago
7 0

Answer:

<h2>30×4=120 ^_^ ...</h2><h2>.......</h2>
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svp [43]
A 80 square meter I’m not quite sure
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14. Translate the following words into an equation and then solve.
Mazyrski [523]

Answer:BRUH

Step-by-step explanation:

BRUH

8 0
3 years ago
Read 2 more answers
What is the slope of a line that passes through (-4,-13) and (19,11)
inysia [295]
The slope of the line is the gradient, which you can find through rise over run

m (gradient) = (y1 - y2) / (x1 - x2)

where (x1, y1) is the coordinate of the first point, and (x2, x2) is the coordinate of the second point

in your question: 
x1 = -4
x2 = 19
y1 = -13
y2 = 11

m = (-13 -11) / (-4 -19) = -24 / -23 = 24/23 or 1.04 (2d.p.)

hope that helps :)
3 0
3 years ago
Read 2 more answers
When a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constan
vitfil [10]

Answer:

a = 6m/s^2

Step-by-step explanation:

Given

When mass = 4kg; Acceleration = 15m/s²

Required

Determine the acceleration when mass = 10kg, provided force is constant;

Represent mass with m and acceleration with a

The question says there's an inverse variation between acceleration and mass; This is represented as thus;

a\ \alpha\ \frac{1}{m}

Convert variation to equality

a = \frac{F}{m}; Where F is the constant of variation (Force)

Make F the subject of formula;

F = ma

When mass = 4kg; Acceleration = 15m/s²

F = 4 * 15

F = 60N

When mass = 10kg; Substitute 60 for Force

F = ma

60 = 10 * a

60 = 10a

Divide both sides by 10

\frac{60}{10} = \frac{10a}{10}

a = 6m/s^2

<em>Hence, the acceleration is </em>a = 6m/s^2<em />

4 0
3 years ago
FIRST ONE TO ANSWER GETS BRAINLIEST!!!!
-Dominant- [34]

Answer:

The maximum distance traveled is 4.73 meters in 0.23 seconds.

Step-by-step explanation:

We have that the distance traveled with respect to time is given by the function,

d(t) = 4.9t^2-2.3t+5.

Now, differentiating this function with respect to time 't', we get,

d'(t)=9.8t-2.3

Equating d'(t) by 0 gives,

9.8t - 2.3 = 0

i.e. 9.8t = 2.3

i.e. t = 0.23 seconds

Substitute this value in d'(t) gives,

d'(t) = 9.8 × 0.23 - 2.3

d'(t) = 2.254 - 2.3

d'(t) = -0.046.

As, d'(t) < 0, we get that the function has the maximum value at t = 0.23 seconds.

Thus, the maximum distance the skateboard can travel is given by,

d(t) = 4.9\times 0.23^2-2.3\times 0.23+5.

i.e. d(t) = 4.9\times 0.0529-2.3\times 0.23+5.

i.e. d(t) = 0.25921-0.529+5.

i.e. d(t) = 0.25921-0.529+5.

i.e. d(t) = 4.73021

Hence, the maximum distance traveled is 4.73 meters in 0.23 seconds.

5 0
3 years ago
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