All categories

3 years ago

(c) If 2a + 7b = 11 and ab = 2, find the value of 4a2 + 49b2.

Mathematics

2 answers:

Mademuasel [1]3 years ago

5 0

I hope this helps you

take both of sides paranthesis square

(2a+7b)^2=(11)^2

4a^2+2.4a.7b+49b^2=121

4a^2+49b^2+56.2=121

4a^2+49b^2=9

kirill115 [55]3 years ago

3 0

Answer:

Step-by-step explanation:

Please see the attached picture for full solution.

You might be interested in

Someone help me pleaseeeee

Rus_ich [418]

Answer:

5. Plane P

6. Line segment DB

7. Line CE

11. Ray BE (with an arrow on top from left to right) and Ray EC (with an arrow on top from left to right)

5 0

3 years ago

Determine if the shape shown is a triangel squear rectangel or hexagon or other

jonny [76]

You can determine the shape if it is a triangle, square, rectangle, or hexagon (or other) by seeing how many sides it has.
a triangle has 3 sides.
a square has 4 sides.
a rectangle also has 4 sides.
a pentagon has 5 sides.
a hexagon has 6 sides.

8 0

3 years ago

Is the absolute value of -8.3, 8.3?

vazorg [7]

Answer:

Step-by-step explanation:

|-8.3| = 8.3

|8.3|= 8.3

5 0

3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

$P = P_{1} + P_{2} + P_{3}$

$P_{1}$ is the probability that all three of them are 13-watt. So:

$P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277$

$P_{2}$ is the probability that all three of them are 18-watt. So:

$P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415$

$P_{3}$ is the probability that all three of them are 23-watt. So:

$P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173$

$P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865$

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

$P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245$

There is a 12.45% probability that one bulb of each type is selected.

3 0

3 years ago

Explain how knowing that 5/8= 0.625 heps you write the decimal for 4 5/8

netineya [11]

To convert a fraction to a decimal , the solution is to divide the numerator by the denominator. Since its a mixed fraction, you already know what the 5/8 of the problem is (0.625). All you do it add the 4 to the decimal and it'll be 4.0625

7 0

4 years ago

(c) If 2a + 7b = 11 and ab = 2, find the value of 4a2 + 49b2.​

(c) If 2a + 7b = 11 and ab = 2, find the value of 4a2 + 49b2.