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maxonik [38]
3 years ago
15

In horse racing, different horses are often assigned different weights to carry. In a particular race, the standard weight carri

ed is 53 kg. The weight carried by each horse cannot differ by more than 4 kg from the standard. What are the maximum and minimum acceptable weights for a horse to carry in this race?
a 49 kg–57 kg
b 53 kg–61 kg
c 51 kg–55kg
d 45 kg–53 kg
Mathematics
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

try c 51 kg -55kg

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The value of -3x y, for x = 2 and y = 3, is -18.<br><br> True<br> False&lt;---- this one
Otrada [13]
I am assuming you are multiplying the y as well. if so, it is -3*2*3 and that would be -18 so it would be true.
7 0
4 years ago
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One cubic centimeter of sand weighs 1.9 grams. Find the amount of sand that the sandcastle bucket can hold.
satela [25.4K]

Answer:

5198.4 grams of sand

Step-by-step explanation:

The bottom cube:

12*12*13= 144*13= 1872

The top pyramid:

144*6= 864

1872+864= 2736 cubic cm

THEN

2736*1.9= 5198.4 g of sand!

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4 0
3 years ago
Knut had picked strawberries during the summer holidays. First days he picked 9090 baskets of strawberries! The next day he was
jasenka [17]

Answer:

190 baskets of strawberries

Step-by-step explanation:

The question is incomplete as it didn't state what we are to determine from the information given.

Let's determine total amount picked in the three days.

1st day picked 90strawberries

Amount of strawberries on 2nd day = 2/3 of the amount picked 1st day

= (2/3)(90)

Amount of strawberries on 2nd day = 60 baskets of strawberries

Amount of strawberries picked on 3rd day = (2/3) of the amount picked 2nd day

= (2/3) (60)

Amount of strawberries picked on 3rd day = 40 basket of strawberries

Total amount picked for 3 days = 90+60+40

= 190 baskets of strawberries

4 0
3 years ago
Plz tell me answer and explain how I could get the answer.
Vinil7 [7]

Answer:   349.5 miles

<u>Step-by-step explanation:</u>

Use the formula distance (d) = rate (r) x time (t):

Since there are two different rates and times so solve each one separately and then add them together.

\large{d_1=r_1\times t_1\quad \rightarrow \quad r=75\ mph, t=150\ minutes}\\\\d_1=\dfrac{75\ miles}{1\ hour}\times \dfrac{1\ hour}{60\ minutes}\times 150\ minutes\\\\d_1=187.5\ miles\\\\\\\large{d_2=r_2\times t_2\quad \rightarrow \quad r=81\ mph, t=2\ hours}\\\\d_2=\dfrac{81\ miles}{1\ hour}\times 2\ hours\\\\d_2=162\ miles\\\\\\\large{\text{Total miles driven = }d_1+d_2}\\\\187.5\ miles +162\ miles = \boxed{349.5\ miles}

7 0
3 years ago
If the outside diameter of a pipe is 1.25 ft and the inside diameter is 0.975 ft how thick is the pipe
Maslowich

Answer:

0.275

Step-by-step explanation:

1.25-0.975=0.275

3 0
3 years ago
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