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3 years ago

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Mathematics

1 answer:

g100num [7]3 years ago

6 0

Answer:

Option B: 3x + 3y - 3

Step-by-step explanation:

In an isosceles triangle, the two congruent sides are equal.

We are told the base is; x - y - 2 units

Now let each of the congruent sides be represented as A.

Thus the perimeter equation will be;

P = 2A + x - y - 2

Now, we are told that the perimeter is; 7x + 5y - 8 units

Thus;

7x + 5y - 8 = 2A + x - y - 2

Rearranging gives;

7x - x + 5y + y - 8 + 2 = 2A

2A = 6x + 6y - 6

Divide through by 2 to give;

A = 3x + 3y - 3 units

You might be interested in

Simiplify. U^2 - 4 / U^2 - 2u

Papessa [141]

Remember
difference of 2 perfect squares
a^2-b^2=(a-b)(a+b)

and the undistributiv propeoryt
ab-ac=a(b-c)

so we have
$\frac{u^{2}-4}{u^{2}-2u}$
undistribute and factor
$\frac{(u-2)(u+2)}{u(u-2)}$
cancel out the one ( (u-2)/(u-2)=1)

$\frac{(u+2)}{u}$ is result

3 0

3 years ago

Read 2 more answers

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$