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2 years ago

What is HL??? HL=__ Image Below

Mathematics

1 answer:

Advocard [28]2 years ago

4 0

Consider a trapezoid GJKF with:

GH=JH (hypothesis)

FL=KL (hypothesis)

so HL is the median of trapezoid GJKF

so: HL=1/2(2.5+1.2)=1.85

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Drag expressions to complete each equation.

irga5000 [103]

Answer:

1st one a^-1 , 2nd one a^b/a^c , 3rd one a^c/b^c

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3 years ago

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Y=3/2x-3

Inga [223]

Answer:

the answer is (2,-6)

Step-by-step explanation:

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3 years ago

Evaluate the surface integral. s x ds, s is the part of the plane 18x + 9y + z = 18 that lies in the first octant.

IceJOKER [234]

In the first octant, the given plane forms a triangle with vertices corresponding to the plane's intercepts along each axis.

$(x,0,0)\implies 18x+9\cdot0+0=18\implies x=1$
$(0,y,0)\implies 18\cdot0+9y+0=18\implies y=2$
$(0,0,z)\implies 18\cdot0+9\cdot0+z=18\implies z=18$

Now that we know the vertices of the surface $\mathcal S$ , we can parameterize it by

$\mathbf s(u,v)=\langle(1-u)(1-v),2u(1-v),18v\rangle$

where $0\le u\le1$ and $0\le v\le1$ . The surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=2\sqrt{406}(1-v)\,\mathrm du\,\mathrm dv$

With respect to our parameterization, we have $x(u,v)=(1-u)(1-v)$ , so the surface integral is

$\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{406}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-u)(1-v)^2\,\mathrm dv\,\mathrm du=\frac{\sqrt{406}}3$

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3 years ago

Which expression is equivalent to (q6)2

Nat2105 [25]

Answer:

q¹²

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Exponential Rule [Powering]: $\displaystyle (b^m)^n = b^{m \cdot n}$

Step-by-step explanation:

Step 1: Define

Identify

(q⁶)²

Step 2: Simplify

Exponential Rule [Powering]: q⁶⁽²⁾
[Exponents] Multiply: q¹²

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3 years ago

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. Using the Binomial Theorem explicitly, give the 15th term in the expansion of (-2x + 1)^19

timofeeve [1]

Let's rewrite the binomial as:
$(1 - 2x)^{19}$

$\text{Binomial expansion:} (1 + x)^{n} = \sum_{r = 0}^n\left(\begin{array}{ccc}n\\r\end{array}\right) (x)^{r}$

Using the binomial expansion, we get:
$\text{Binomial expansion: } (1 - 2x)^{19} = \sum_{r = 0}^{19}\left(\begin{array}{ccc}19\\r\end{array}\right) (-2x)^{r}$

For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.

Thus, the fifteenth term is:
$\text{Binomial expansion (15th term):} \left(\begin{array}{ccc}19\\14\end{array}\right) (-2x)^{14}$

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3 years ago