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3 years ago

8

An isosceles triangle with base 12mm and perimeter 28mm

1 answer:

olga55 [171]3 years ago

5 0

Answer:

Side lengths of 12,8,8

Step-by-step explanation:

In an isosceles triangle the base is different from the other 2 sides which are the same so

28=12+2x (x being the side of the repeat sides)

14=6+x (divide both sides by 2)

8=x (subtract 6 from both sides) so

Side lengths of 12,8,8

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3 years ago

A city planner is rerouting traffic in order to work on a stretch of road. The equation of the path of the old route can be desc

earnstyle [38]

Answer: D

Explanation:

The equation of a line in the point slope form is expressed as

y - y1 = m(x - x1)

where

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y = 2x/5 - 4

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y = mx + c

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1 year ago

Vesna [10]

nA.Replace ''the difference with subtraction

B.Replace''the number of serving '' with a variable x.

C.Replace ''four'' with 4.

Step-by-step explanation: .

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1 year ago

Can please someone help me? The question is in the picture. Thank you

maxonik [38]

Answer:

C. $9\pi$

Step-by-step explanation:

We proceed to solve for $x$ by the Pythagorean Theorem:

$(x+1)^{2} = (x-1)^{2} + (2\cdot x - 4)^{2}$ (1)

$x^{2} + 2\cdot x + 1 = (x^{2}-2\cdot x + 1) + (4\cdot x^{2}-16\cdot x + 16)$

$x^{2} + 2\cdot x + 1 = 5\cdot x^{2}-18\cdot x + 17$

$4\cdot x^{2} -20\cdot x + 16 = 0$

$(2\cdot x)^{2} -10\cdot (2\cdot x) + 16 = 0$

$(2\cdot x -8)\cdot (2\cdot x -2) = 0$

There are two different solutions:

$x_{1} = 4$ , $x_{2} = 1$

A solution of $x$ is considered realistic when the length of every side of the triangle is a not negative number.

$x_{1} = 4$

$UV = 2\cdot (4) - 4$

$UV = 4$

$WU = 4 - 1$

$WU = 3$

$WV = 4 + 1$

$WV = 5$

$x_{2} = 1$

$UV = 2\cdot (1) - 4$

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$WV = 2$

The only realistic set of solutions is for $x = 4$ , and the radius of the circle is represented by the line segment $WU$ : $r = 3$

And the area of the circle ( $A$ ) is calculated from the following formula:

$A = \pi\cdot r^{2}$

$A = \pi\cdot (9)$

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The correct answer is C.

6 0

3 years ago