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inessss [21]
3 years ago
7

29 mm 120 mm What is the perimeter? millimeters

Mathematics
1 answer:
Svetlanka [38]3 years ago
6 0

Answer:

I'm assuming it's a rectangle because you only specified two sides. If so, the perimeter is 298 millimeters.

Step-by-step explanation:

Add the two numbers up, 29 + 120 = 149, and multiply it by 2. You get 298 millimeters.

Another way is to add 29 + 29 + 120 + 120 = 298 millimeters.

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Using Cramer’s Rule, what is the value of x in the system of linear equations below?
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Answer:

x = 8 is the answer

Step-by-step explanation:

In order to find  the solution of Equations using Cramer's rule

we shall find  matrices

D= \left[\begin{array}{ccc}5&-3\\1&-2\end{array}\right]=

D_{1} = \left[\begin{array}{ccc}4&-3\\-16&-2\end{array}\right]

l D l  = Determinant formed by coefficient  of variables

Here l D l = 5X(-2) - 1X(-3)=  -10+3 = -7

l D1 l =   Determinant formed by replacing first column In D by numbers on the right side of the equations

Here  l D1 l = 4X(-2)- (-16)X(-3)= -8-48 = -56

here x is  given by \frac{l D1 l}{l D l}

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3 0
3 years ago
There are 30 students in Ms. Prema's history class. 18 of these students turned in an extra credit project at the end of the sem
kow [346]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Using only definition 2.2.3, prove that if (x
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7 0
3 years ago
50 POINTS PLEASE HELP
lorasvet [3.4K]

Answer:

perimeter= 5\sqrt{15} + 5\sqrt{5}

Step-by-step explanation:

Given ,

A right angle triangle ΔABC with right angle at C and ∠A=30° and AC=5√5 units .

Let ∠A=A,∠B=B∠C=C and AC=b,BC=a,CA=b .

Implies perimeter of triangle = a+b+c .

now

tanA=\frac{a}{b} \\\a=b*tanA\\a= 5\sqrt{5} *tan30^0\\a=\frac{5\sqrt{5}}{\sqrt{3}}

and

cosA=\frac{b}{c} \\\c=\frac{b}{cosA} \\c=\frac{10\sqrt{5}}{\sqrt{3} }

implies ,

perimeter = \frac{5\sqrt{5}}{\sqrt{3}} + 5\sqrt{5} +\frac{10\sqrt{5}}{\sqrt{3} }\\perimeter= 5\sqrt{15} + 5\sqrt{5}

8 0
3 years ago
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ale4655 [162]

Answer:

2:1

Step-by-step explanation:

12 divided by 6 = 2

6 divided by 6 = 1

2:1

7 0
3 years ago
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