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Kitty [74]
3 years ago
14

The daily recommended allowance of vitamin C for six graders is 45 MG 1 orange has an about 75% of the recommended daily allowan

ce of five and see how many milligrams are in 150 guess I consider using the double number line

Mathematics
1 answer:
slamgirl [31]3 years ago
8 0

Answer:

So there are 33.75 MG in an orange.

Step-by-step explanation:

I think your full question in the photo below,

Here is my answer:

1 orange has an about 75% of the recommended daily allowance: 0.75*45 = 33.75 MG.

So there are 33.75 MG in an orange.

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Angles 6, 4, and 2 are congruent to angle 8.

Step-by-step explanation:

By the Opposite Angles Theorem, angle 6 is congruent to angle 8.

By the Corresponding Angles Theorem, angle 4 is congruent to angle 8.

By the Corresponding Angles Theorem, angle 2 is congruent to angle 6, and angle 6 is congruent to angle 8, so angle 2 is congruent to angle 8 by the Transitive Property.

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The exponent indicates how many times the ......... is used as a factor
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Base  or base number
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What does the left side of the equal<br> sign simplify to?
polet [3.4K]

Answer:

Step-by-step explanation:

An expression is a mathematical statement that does not contain an equal sign. It cannot be solved for unless the value of the variable is given. An equation is a mathematical statement or sentence comprised of two equalities or expressions joined with an equal sign.

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3 years ago
Read 2 more answers
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

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3 years ago
NEED HELP M&lt;1 <br><br> Use the given information
kondor19780726 [428]
The line should equal 180 degrees, so we would subtract 122 from 180 to get 58 degrees. Since the triangle is equilateral, each angle is the same measure, so I believe m<1 should be 58 degrees.
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