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Marina86 [1]
3 years ago
9

The pineapple is seven times as heavy as an orange or pineapple also weighs 870 g videos in the orange what is the total weight

in grams for the pineapple and orange
Mathematics
2 answers:
borishaifa [10]3 years ago
8 0
The total is 6960grams
kirill [66]3 years ago
5 0
Okay so what you want to do it's multiply the 7 by 870 which gives you the answer you were given above but remember problems like these you need to multiply and make sure to label your units MY teachers used to tell me no naked number so make sure everything is labeled and I hope you can understand what I'm saying because I know it might be hard but I'm just a Stanley trying to help so idk I'm kinda going crazy and yapping about myself so why don't I just tell u my whole life story I grew up in a house lol jk well I hoped this helped bye
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X - Y = 11<br> 2x + Y = 19<br><br> where do the lines cross
katen-ka-za [31]
Just solve the equations:
<span>A) X - Y = 11
B) 2x + Y = 19
Multiply A) by -2
</span>
<span>A) -2X  +2Y = -22 then add to B)
</span><span>B) 2x + Y = 19
3Y = -3
Y=-1
</span>
<span>A) X - -1 = 11
x = 10

</span>
4 0
3 years ago
Elaina Kicks A Soccer Ball With An Initial Velocity Of
VladimirAG [237]

By using the motion equations, we conclude that she will make the goal.

<h3>Will she make the goal?</h3>

First, we know that the goal is 20ft away, and the position in x is given by:

x = 28*cos(58)*t

Let's find the time in which the ball will travel these 20 ft.

20 =  28*cos(58)*t

(20)/(28*cos(58)) = t = 1.35

So the horizontal distance is covered in 1.35 seconds, now let's see which is the height of the ball at that time.

The height equation is:

y = -16*t^2 +28*sin(58)*t

Evaluating in t = 1.35 we get:

y = -16*(1.35)^2 +28*sin(58)*1.35\\\\y = 2.9

And the goal is 5ft tall, so we can conclude that she will make the goal.

If you want to learn more about motion equations:

brainly.com/question/19365526

#SPJ1

5 0
2 years ago
What size bag of fertilizer should the jamisons buy?
vredina [299]
Where is the rest of the question
5 0
3 years ago
One pumpkin weighs 7 lb 12 oz. A second pumpkin with 10 pounds 4 ounces. A pumpkin weighs 2 pounds 9 ounces more than the second
Nesterboy [21]

Answer:  30 lbs, 13 oz

This converts to 493 ounces

===============================================================

Explanation:

Pumpkin A weighs 7 lb, 12 oz. Let's convert that to ounces only

1 lb = 16 oz

7 lbs = 112 oz (multiply both sides by 7)

7 lbs, 12 oz = 112 oz + 12 oz .... add 12 oz to both sides

7 lbs, 12 oz = 124 oz

Pumpkin A's weight of 7 lbs, 12 oz is the same as 124 oz

-----------

Do something similar for pumpkin B

1 lb = 16 oz

10 lbs = 160 oz

10 lbs, 4 oz = 160 oz + 4 oz

10 lbs, 4 oz = 164 oz

Pumpkin B weighs 164 oz

-----------

Let's convert the "2 lbs, 9 oz" to ounces only

1 lb = 16 oz

2 lbs = 32 oz

2 lbs, 9 oz = 32 oz + 9 oz

2 lbs, 9 oz = 41 oz

So pumpkin C is 41 ounces heavier compared to pumpkin B

The weight of pumpkin C is 164+41 = 205 ounces

----------

We have these three weights in ounces only

  • Pumpkin A = 124 oz
  • Pumpkin B = 164 oz
  • Pumpkin C = 205 oz

Adding those three ounces only weights leads to

124+164+205 = 493 ounces

The total weight of all three pumpkins is 493 ounces

Now let's convert back to the unit "pounds, ounces"

Divide that result over 16 since there are 16 ounces per pound

493/16 = 30.8125

Ignore the decimal portion. The 30 is the quotient which means 493-30*16 = 13 is the remainder.

In other words,

493/16 = 30 remainder 13

which leads to

493 ounces = 30 pounds, 13 ounces

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

As a check,

1 lb = 16 oz

30 lbs = 480 oz

30 lbs, 13 oz = 480 oz + 13 oz

30 lbs, 13 oz = 493 oz

The answer is confirmed.

8 0
3 years ago
Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
2 years ago
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