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3 years ago

4

Using the GCF you found in Part B, rewrite 72 + 81 as two factors. One factor is the GCF and the other is the sum of two numbers

that do not have a common factor. Show your work. (4 points)

1 answer:

zloy xaker [14]3 years ago

5 0

Answer:- 72 + 81 = 9 * 17:-

Find the GCF of 72 and 81 by first writing out the prime factors:-

72 = 2*2*2*3*3

81 = 3*3*3

Picking out common factors we get GCF = 3*3 = 9

So 72 + 81 = 9(8 + 9) (using the distributive law)

= 9 * 17

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Answer:

5

Step-by-step explanation:

just look at the graph

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3 years ago

Calculate the following without the use of a calculator /

11111nata11111 [884]

Answer:

5

Step-by-step explanation:

6 - (-4) - 5 =

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3 years ago

What's the answer for this problem number 6 and how did you get your answer?

lorasvet [3.4K]

$\frac{7}{b + 5} = \frac{10}{5} \\$

$\frac{7}{b + 5} = 2 \\$

Multiply the sides of the equation by ( b + 5 ) :

$7 =2 \times (b + 5)$

$7 = 2b + 10$

Subtract the sides of the equation minus 10 :

$7 - 10 = 2b$

$- 3 = 2b$

Divided the sides of the equation by 2 :

$b = - \frac{3}{2} \\$

____________________________

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4 0

4 years ago

From a bowl containing five red, three white, and seven blue chips, select four at random and without replacement. Compute the c

snow_lady [41]

Answer:

The probability is $\frac{5}{9}$

Step-by-step explanation:

Let A be the event of one red, zero white, and three blue chips,

And, B is the event of at least three blue chips,

Since, A ∩ B = A (because If A happens that it is obvious that B will happen )

Thus, the conditional probability of A if B is given,

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$

Now, red chips = 5,

White chips = 3,

Blue chips = 7,

Total chips = 5 + 3 + 7 = 15

Since, the probability of one red, zero white, and three blue chips, when four chips are chosen,

$P(A)=\frac{^5C_1\times ^3C_0\times ^7C_3}{^{15}C_4}$

$=\frac{5\times 35}{1365}$

$=\frac{175}{1365}$

$=\frac{5}{39}$

While, the probability that of at least three blue chips,

$P(B)=\frac{^8C_1\times ^7C_3+^8C_0\times ^7C_4}{^{15}C_4}$

$=\frac{8\times 35+35}{1365}$

$=\frac{315}{1365}$

$=\frac{3}{13}$

Hence, the required conditional probability would be,

$P(\frac{A}{B})=\frac{5/39}{3/13}$

$=\frac{65}{117}$

$=\frac{5}{9}$

4 0

3 years ago

In the figure, angle 4 is an exterior angle to angle AUL.

Whitepunk [10]

The measure of angle 4 is equal to the sum of the measures of the two nonadjacent interior angles because the angle 4 is exterior to the given interior angles.

<h3>Exterior angles of a triangle</h3>

The sum of interior angles is equal to the exterior. From the given figure;

Interior angles are 25 degrees and 52 degrees.

The measure of angle 4 is equal to the sum of the measures of the two nonadjacent interior angles because the angle 4 is exterior to the given interior angles.

2) <4 = 25 + 52

<4 = 77 degrees

Hence the measure of angle 4 is 77 degrees

Learn more on exterior angles here: brainly.com/question/2546141

#SPJ1

6 0

2 years ago