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bija089 [108]
3 years ago
12

Given a=6 b=12 c=3 Can you construct a triangle?​

Mathematics
2 answers:
12345 [234]3 years ago
8 0

Answer:

no because c, the hypotenuse, should always be larger than a and b added

Step-by-step explanation:

Greeley [361]3 years ago
8 0

c=3 You can't construct a triangle.

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which of the following formulas would find the lateral area of a right cylinder where h is the height and r is the radius
vlada-n [284]

Answer:

B. LA=2\pi rh

Step-by-step explanation:

The lateral area of the right cylinder refers to the curved surface area.

The lateral area of the right cylinder does not include the two circular bases.

The lateral area is given by the formula;

LA=2\pi rh

The correct choice is B.

6 0
3 years ago
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Patrick ate 2/3 of a fruit bar write an equivalent fraction for the amount of fruit bar Patrick ate
Marina86 [1]
To get an equivalent fraction, multiply the fraction by a whole number. For example, you can do 2/3 times 2 to both sides which is 4/6.
5 0
3 years ago
What are the answers plz help me
Serga [27]

For number 3 it would be |-10| and |10|.

Number 5a is supposed to <, because -1 is greater than -7 when you look at the line graph. The same goes for 5b. 5d is >, because 0 is always greater than -1 and it shows that on the line graph that you have there. 5e is =.

I don't see anything else wrong though. Just the ones I listed.

Hope that helps!

8 0
3 years ago
Identify a pair of factors of -35 that has a sum of -2​
lions [1.4K]

Answer:

-7 and 5

Step-by-step explanation:

-7*5= -35

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5 0
3 years ago
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Examine the system of equations.
MrMuchimi

answer:

x =  - 2 \\ y = 2 \\  \\ ( - 2, \: 2)

explanation:

y =  -  \frac{1}{2} x + 1 \\ y = 2x + 6 \\ first \: equation \: add \:  \frac{1}{2} from \: both  \\  sides \\   \\ second \: equation \: subtract \: 2x \: from  \\  both \: sides \\  \\ y +  \frac{1}{2} x = 1 \\ y - 2x = 6 \\ solve \: the \: first \: equation \\  \\ y +  \frac{1}{2} x = 1 \\ subtract \:  \frac{x}{2} from \: both \: sides \\  \\ y =  -  \frac{1}{2} x + 1 \\ substitute \:  -  \frac{x}{2}  + 1 \: for \: y \: in \: the  \\ other \: equation \\  \\  -  \frac{1}{2} x + 1 - 2x = 6 \\ add \:   \frac{1}{2} x \: to \: 2x \\  \\  -  \frac{5}{2} x + 1 = 6 \\ subtract \: 1 \: from \: both \: sides \\  \\  -  \frac{5}{2} x = 5 \\ divide \: both \: sides \: by \:  -  \frac{5}{2}  \\  \\ x =  - 2 \\ substitute \: the \: value \: of \: x \: into \\ an \: equation \\  \\ y =  -  \frac{1}{2} ( - 2) + 1 \\ distribute \\  \\ y = 1 + 1 \\ add  \\  \\ y = 2 \\  \\ ( - 2, \: 2)

7 0
3 years ago
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