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3 years ago

2 2/7× 11/12 what is the ans

Mathematics

2 answers:

iren [92.7K]3 years ago

8 0

The answer to your question is 44/21 or 2.09524

lisabon 2012 [21]3 years ago

3 0

Answer:

There are 3 different forms of the answer.

Step-by-step explanation:

1. 44/21

2. 2.10

3. 2 2/21

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What percent of the figure is shaded

GREYUIT [131]

Answer:

a recentangle?

Step-by-step explanation:

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3 years ago

Which of the following best classifies a quadrilateral with coordinates A(2,6), B(5, 1), C(10, 4), and D(7,9)?

navik [9.2K]

Answer:

a rhombus

Step-by-step explanation:

If you graph the problem, you can see the shape of the quadrilateral. I attached a picture of a graph below. I hope this helped you!! Have a great rest of your day.

3 0

3 years ago

What is the vertex of y = x2 + 2x +2 O (1,-1) O (2,-2) O (1, 2) 0 (-1, 1)

fredd [130]

Answer:

vertex = (- 1, 1 )

Step-by-step explanation:

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - $\frac{b}{2a}$

y = x² + 2x + 2 ← is in standard form

with a = 1 and b = 2 , then

x = - $\frac{2}{2}$ = - 1

substitute x = - 1 into the equation for y- coordinate of vertex

y = (- 1)² + 2(- 1) + 2 = 1 - 2 + 2 = 1

vertex = (- 1, 1 )

8 0

3 years ago

Suppose that a water fountain produces a stream of water that follows the shape of a parabola. Assigning the origin of a coordin

irina1246 [14]

Vertex = (6, 5)
axis of symmetry is line x = 6
A third point in the parabola is (12, 0)

equation using the vertex is y = -a(x - 6)^2 + 5 = -a(x^2 - 12x + 36) + 5 = -ax^2 + 12ax - 36a + 5
since the parabola passes through point (0, 0)
i.e. -36a + 5 = 0
a = 5/36 = 0.139

Therefore, the equation is y = -0.139x^2 + 12(0.139)x = -0.139x^2 + 1.667x

7 0

4 years ago

CALC- limits please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago