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3 years ago

The two parallelograms in the figure are similar. What is the value of x?

Mathematics

2 answers:

puteri [66]3 years ago

5 0

Answer:

The answer is D, 30.4

Step-by-step explanation:

Because the two parallelograms are similar, the relationship between the sides of them are directly proportional.

So:

$\frac{9}{8} =\frac{x}{27}$

Cross multiply

8x = 243

÷8 both sides

x = 30.375 ≈ 30.4

The answer is D, 30.4

12345 [234]3 years ago

3 0

ANSWER

D. 30.4

EXPLANATION

The two parallelograms are similar, therefore the corresponding sides must be in the same proportion:

$\frac{x}{9} = \frac{27}{8}$

Multiply both sides of the equation by 8. This implies that,

$x = \frac{27}{8} \times 9$

$x = 30.375$

Rounding to the nearest tenth gives us:

$x = 30.4 \: units$

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Fiona, Dan and Stephen share £66 in a ratio 1:3:2. How much money does each person get?

Viefleur [7K]

Answer:

Fiona=11

Dan=33

Stephen=22

Step-by-step explanation:

Fiona:1/2+3+1×66

1/6×66

=11

Dan: 3/6×66=33

Stephen:2/6×66=22

8 0

3 years ago

Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla

Ahat [919]

(i) Yes. Simplify $f(x,y)$ .

$\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}$

Now compute the limit by converting to polar coordinates.

$\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0$

This tells us

$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1$

so we can define $f(0,0)=1$ to make the function continuous at the origin.

Alternatively, we have

$\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2$

and

$\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2$

Now,

$\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0$

$\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0$

so by the squeeze theorem,

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and $f(x,y)$ approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

$\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x$

$f(0,y)$ and $f(x,0)$ are undefined, so there is no way to make $f(x,y)$ continuous at (0, 0).

(iii) No. Similarly,

$\dfrac{x^2 + y}y = \dfrac{x^2}y + 1$

is undefined when $y=0$ .

5 0

1 year ago

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xxMikexx [17]

Answer:

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