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Eddi Din [679]
4 years ago
12

PLEASE HELP I"M ON A TIME LIMT!!

Mathematics
1 answer:
NemiM [27]4 years ago
8 0

Answer: C) 1560

Step-by-step explanation:

So the first part is splitting the rectangle from the triangle so the area for the rectangle is 1440 (40 x 36) and the area for the triangle is 120 ((6x40)/ 2)

when you add them up the equal 1560 (1440 + 120)

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Circle the problem in wich you need to regroup. Use the strategy that is easier to find the sum. A) 496+284. B) 482+506
Musya8 [376]
B, because I’m A, it has no number to regroup from(all the numbers can go into each other when yo add) In B, 6 can’t go into 2, so you’ll need to regroup, - hopefully this helped and hopefully I get the answer correct
5 0
3 years ago
Use fraction strips to find the difference. Write your answer in simplest form. 1/2-1/5
zzz [600]

Answer:

It can not be simplified after 3/10 so it stays that way

Step-by-step explanation:

1/2=5/10

1/5=2/10

5/10-2/10=3/10

3 0
3 years ago
The smallest object visible with your eyes is similar to the width of a piece of hair, which is 1×10−4 meters wide. Using an opt
Sloan [31]

Answer:

B. 5\times10^{2}

Step-by-step explanation:

We are told that the smallest object visible with our eyes is similar to the width of a piece of hair, which is 1\times 10^{-4} meters wide.

Using an optical microscope, we can see items up to 2\times 10^{-7} meters wide.

To find the objects we can see with our eyes are how much larger than the objects we can see with an optical microscope, we can set an equation as:

\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1*10^{-4}}{2*10^{-7}}

Using the exponent rule of quotient \frac{a^m}{a^n}=a^{m-n} we will get,

\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=\frac{1}{2}*10^{-4-(-7)}

\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{-4+7}

\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10^{3}

\frac{\text{The width of the object we can see with our eyes}}{\text{The width of the objects we can see with microscope}}=0.5*10\times 10^{3-1}

\text{The object we can see with our eyes}=5\times10^{2}*\text{The objects we can see with microscope}

Therefore, the objects we can see with our eyes are 5\times10^{2} times larger than the objects we can see with an optical microscope and option B is the correct choice.

3 0
3 years ago
Find the missing length of the missing side. If necessary, round to the nearest tenth
sladkih [1.3K]
A^2+b^2=c^2
there for 5^2+14^2=221
then you take the sqr root of 221 to find c
the square root of 221 is 14.86606875
which rounds to 14.9
so the answer is D
8 0
3 years ago
Help I need this done today !!!!
Vsevolod [243]

Answer:

c

Step-by-step explanation:

To find the maximum point, differentiate:

D' = -8p +160

Set to zero and solve for p:

D' = 0

⇒ -8p +160 = 0

⇒ 8p = 160

⇒ p = 20

Input p = 20 into the original equation to find D:

D = -4(20)² + 160(20) -310 = 1290

Therefore, the price is $20 and the maximum number of drills is 1,290

6 0
2 years ago
Read 2 more answers
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