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Alisiya [41]
3 years ago
15

The base of a triangle is 21 inches and the height is 12 inches which of these expressions correctly shows how to calculate the

area of a triangle
a (21 x 12) divided by 2
b ( 21 + 12) x 2
c (21 + 12) divided by 2
d ( 21 x 12) x 2
Mathematics
1 answer:
I am Lyosha [343]3 years ago
5 0

½ x 21 x 12=126

So the answer is a

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The parent function of the function g(x) = (x – h)2 + k is f(x) = x2. The vertex of the function g(x) is located
Katen [24]

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Answer:

  (h, k) = (9, -8)

Step-by-step explanation:

The function g(x) is "vertex form" for a quadratic whose vertex is (h, k). We are told the vertex is (9, -8), so we have ...

  (h, k) = (9, -8) . . . . the values of h and k

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Determine if the equation is parallel, perpendicular, or neither;
USPshnik [31]
The equations are parallel because the slopes are opposite reciprocals, which means if you flip -3/2 and multiply by -1 you get 2/3.
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4 years ago
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What is the gcf of 18 and 54​
katrin2010 [14]

Answer:

18 is the GCF

Step-by-step explanation:

Factors of 18: 1, 2, 3, 6, 9, 18

Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54

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A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the
Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= \pi r^2h

According to the problem,

\pi r^2 h=25

\Rightarrow h=\frac{25}{\pi r^2}

The surface area of the base of the can is = \pi r^2

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× \pi r^2)

The lateral surface area of the can is = 2\pi rh

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× 2\pi rh)

                                                 =\$2.5 \pi r h

Total cost of metal is C= 2.00 \pi r^2+2.5 \pi r h

Putting h=\frac{25}{\pi r^2}

\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}

\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}

Differentiating with respect to r

C'=4\pi r- \frac{62.5}{ r^2}

Again differentiating with respect to r

C''=4\pi + \frac{125}{ r^3}

To find the minimize cost, we set C'=0

4\pi r- \frac{62.5}{ r^2}=0

\Rightarrow 4\pi r=\frac{62.5}{ r^2}

\Rightarrow  r^3=\frac{62.5}{ 4\pi}

⇒r=1.71

Now,

\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0

When r=1.71 cm, the metal cost will be minimum.

Therefore,

h=\frac{25}{\pi\times 1.71^2}

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

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
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