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galben [10]
2 years ago
10

I forgot the unit rate of $6.25 for 5lbs. of apples

Mathematics
1 answer:
VladimirAG [237]2 years ago
5 0
$6.25 is the cost of 5 lbs. of apples. Here the question is 'what is the cost of 1 lb of apple?' To find the unit rate or the cost of 1 lb of apple, divide the sum by the given quantity. The sum is @6.25 and the given quantity is 5 lbs. So the unit rate = 6.25/5 = $1.25
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Dibuje la gráfica de la ecuación 2x + 3y = 6
Llana [10]

Answer:

y=3-3/2

Step-by-step explanation:

8 0
2 years ago
Simplify the following expression: (5x2 + 3x + 4) − (2x2 − 6x + 3). If the final answer is written in the form Ax2 + Bx + C, wha
Ratling [72]
Combine like terms. remember to keep the signs.


5x² - 2x² = 3x²
3x - (-6x) = 3x + 6x = 9x
4 - (-3) = 7

3x² + 9x + 7 = Ax² + Bx + C

9 = B, because 9 is in the B's place

9 is your answer

hope this helps
6 0
3 years ago
Read 2 more answers
The measure of angle B is 65.8°. find the measure of the complement and supplement of angle B.​
viktelen [127]

Answer:

The fourth choice

24.2° and 114.2°

Step-by-step explanation:

Complementary angles add up to 90°

Supplementary angles add up to 180°

*a trick to remember this is to know that C is before S in the alphabet, and 90 is before 180, so Complementary is 90 and Supplementary is 180

Step 1:  Find the complementary angle

  Take 90° and subtract the give angle to find the angle you would need...

       90° - 65.8° = 24.2°

Step 1:  Find the supplementary angle

  Take 180° and subtract the give angle to find the angle you would need...

       180° - 65.8° =  114.2°

3 0
3 years ago
A cannonball of mass 1kg is shot vertically upward from the top of a building with an unknown velocity v_0(m/sec).v 0 ​ (m/sec).
Bumek [7]

Taking the upward direction to be positive, the cannonball's height y(t) in the air at time t is given by

y(t)=y_0+v_0 t-\dfrac g2t^2

where g is the magnitude of the acceleration due to gravity, 10 m/s^2, and y_0 is the height of the building from which the ball is being thrown.

At the moment the cannonball reaches its maximum height of 30 m, its velocity at that time is 0, so that

0^2-{v_0}^2=-2g(30\,\mathrm m-y_0)\implies v_0=\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}

Substitute this into the height equation above, and let t=2\,\mathrm s, for which we have y(2\,\mathrm s)=30\,\mathrm m:

30\,\mathrm m=y_0+\sqrt{\left(20\dfrac{\rm m}{\mathrm s^2}\right)(30\,\mathrm m-y_0)}(2\,\mathrm s)-\left(5\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)^2

Solve for y_0: (units omitted for brevity; we know that y_0 should be given in m)

30=y_0+4\sqrt{150-5y_0}-20

50-y_0=4\sqrt{150-5y_0}

(50-y_0)^2=\left(4\sqrt{150-5y_0}\right)^2

2500-100y_0+{y_0}^2=16(150-5y_0)

{y_0}^2-20y_0+100=0

(y_0-10)^2=0

\implies\boxed{y_0=10\,\mathrm m}

3 0
3 years ago
- 21xº<br> What is the answer
Sav [38]

Answer:

The answer to ur question is -1

6 0
3 years ago
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