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3 years ago

The area of the triangle is 28.7 square centimeters. What is the base length of the triangle in centimeters?

Mathematics

1 answer:

Montano1993 [528]3 years ago

3 0

In order to derive the base of a triangle from its area, you need its height as well.

In fact, if we solve the area formula for the base, we have

$A=\dfrac{bh}{2} \iff 2A = bh \iff b=\dfrac{2a}{h}$

So, the base length would be

$b=\dfrac{2\cdot 28.7}{h} = \dfrac{57.4}{h}$

where h is the height relative to the base you're interested in.

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Prove that: (12^13–12^12+12^11)(11^9–11^8+11^7) is divisible by 3, 7, 19, and 37. The answer should be like: x^b*3*7*19*37. Also

n200080 [17]

Answer:

Step-by-step explanation:

Given (12^13–12^12+12^11)(11^9–11^8+11^7),

(12^13–12^12+12^11)(11^9–11^8+11^7) =

[(12^12)12 – 12^12 + 12^11][(11^8)11 – 11^8 + 11^7)

[(12^12)(12 – 1) + 12^11][(11^8)(11 – 1) + 11^7] =

(12^12(11) + 12^11)(11^8(10) + 11^7) =

(12^11(12x11) + 12^11)(11^7(11x10) + 11^7) =

[(12^11)(12x11 + 1)][(11^7)(11x10 + 1)] =

[(12^11)x(11^7)](12x11 + 1)(11x10 + 1) =

[(12^11)x(11^7)](133 x 111) =

[(12^11)x(11^7)](14763) =

[(12^11)x(11^7)](3x7x19x37)

From here, it is clear that the given number is divisible by 3, 7, 19 and 37.

6 0

3 years ago

In factored form please help

Brilliant_brown [7]

Answer:

3rd option

Step-by-step explanation:

$\frac{3x^2-3}{x^2-5x+4}$ ( factorise numerator and denominator )

3x² - 3 ← factor out 3 from each term

= 3(x² - 1²) ← x² - 1 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b)

x² - 1

= x² - 1²

= (x - 1)(x + 1) , then

3x² - 3 = 3²(x - 1)(x + 1) ← in factored form

--------------------------------

x² - 5x + 4

consider the factors of the constant term (+ 4) which sum to give the coefficient of the x- term (- 5)

the factors are - 1 and - 4 , since

- 1 × - 4 = + 4 and - 1 - 4 = - 5 , then

x² - 5x + 4 = (x - 1)(x - 4)

then

$\frac{3x^2-3}{x^2-5x+4}$ = $\frac{3(x-1)(x+1)}{(x-1)(x-4)}$ ← in factored form

3 0

2 years ago

Nicole and her grandmother lives in the same neighborhood,but a forest of trees separate their house.if you left Nicole's house

Masteriza [31]

Wouldn't that be 3 miles?

6 0

3 years ago

Read 2 more answers

Anesh normally buys 1.5 litre bottles of soft drink. At the supermarket he notices a promotion bottle that has 25 percent extra.

slavikrds [6]

Answer:

6ml

Step-by-step explanation:

=25%÷100%

=0•25

1.5l to ml

1L=1000ml

=1.5000ml÷0.25

=6ml

6 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago