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2 years ago

Simplify (4x) * mn * 15 * 2

1 answer:

8 0

Since it's multiplication we can Multiply the constants. In this case 4, 15, and 2. We get 120. We now then multiply the variables. We then get the answe 120mnx

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11th grade geometry:

Aliun [14]

Answer:

The perimeter is 72 units and the area is 149 square units.

Step-by-step explanation:

$\triangle SBA$ has coordinates $S(15,-8),B(-2,21)$ and $A(0,0)$

Using the distance formula.........

Length of side $SB = \sqrt{(15+2)^2+(-8-21)^2}= \sqrt{17^2+(-29)^2}= \sqrt{1130}$

Length of side $BA= \sqrt{(-2)^2+(21)^2}= \sqrt{445}$

Length of side $AS =\sqrt{(15)^2+(-8)^2}=\sqrt{289}=17$

So, the perimeter of the triangle will be: $(SB+BA+AS)= \sqrt{1130}+ \sqrt{445}+17 =71.71... \approx 72$ units. (Rounded to the nearest unit)

The height of the triangle for the corresponding base $SB$ is 8.89 units.

Formula for the Area of triangle, $A= \frac{1}{2}(base\times height)$

So, the area of the $\triangle SBA$ will be: $\frac{1}{2}(\sqrt{1130}\times 8.89)= 149.42... \approx 149$ square units. (Rounded to the nearest unit)

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3 years ago

Miguel's family drinks 3 gallons of milk

Luden [163]

D because ugh I don’t feel like explaining

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2 years ago

Graph the circle x2+y2-10x-10y+25=0

ExtremeBDS [4]

Answer:

x2+y2-10x-10y+25=0

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3 years ago

A triangle with sides 11m , 13m and 18m is a right triangle. A True B False

Julli [10]

$\bold{\huge{\pink{\underline{ Solution}}}}$

$\bold{\underline{ Given :-}}$

A triangle with sides 11m, 13m and 18m

$\bold{\underline{ To \: Find :-}}$

We have to check it whether it is right angled triangle or not?

$\bold{\underline{ Let's \: Begin :-}}$

$\sf{\red{ In \:right \:angled\ : triangle, }}$

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

We imply it in the given triangle ,

$\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}$