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Vesna [10]
3 years ago
11

Consider the following statements.

Mathematics
1 answer:
Slav-nsk [51]3 years ago
5 0

Answer:

.

Step-by-step explanation:

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Evaluate b^2 + 5a – 20
Rina8888 [55]

Answer: it cant even be factored

Step-by-step explanation:

7 0
2 years ago
Mr.bryce spends 712$ each month on rent.Mr.brewter spends 845$ each month on rent.How much more money does Mr. Brewster spend on
Talja [164]
Easy

Subtract.

$845 - $712 = 133
6 0
3 years ago
I'll give 100 points to whoever answers​
ollegr [7]
The answer is 80%
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7 0
3 years ago
Read 2 more answers
How many different ways are there to choose a dozen donuts from the 4 varities at a donut shop so that you get atleast one donut
Leno4ka [110]

There are 330 different ways for choosing a dozen donuts from the 4 varieties at a donut shop. (at least one donut of every variety must be selected)

<h3>How to calculate the number of ways to select items?</h3>

There are 'r' items from 'n' different varieties (repetition allowed). Then, the number of ways to select items is given by

The number of ways =  _{n+r-1}C_r

Where the combination _{n}C_r is calculated as

_{n}C_r=\frac{n!}{r!(n-r)!}

<h3 /><h3>Calculation:</h3>

It is given that there are 4 varieties of donuts in a shop. I.e., n = 4

Number of donuts to be selected r = 12 (one dozen)

And also given that at least one donut of every variety has to be selected.

Since there are 4 varieties, at least one from each of these means the count is 4.

So, the remaining number of donuts to be selected is 12 - 4 = 8.

So, r becomes 8 i.e., r = 8

On substituting,

the number of ways of selecting the remaining 8 donuts = _{4+8-1}C_4

⇒ _{11}C_4

⇒ \frac{11!}{4!(11-4)!}

⇒ \frac{11!}{(4!)(7!)}

⇒ 330

Therefore, there are 330 different ways for choosing a dozen donuts from the 4 varieties at a donut shop.

Learn more about combinations here:

brainly.com/question/11732255

#SPJ4

5 0
2 years ago
find the value of "a" and "b" for which the limit exists both as x approaches 1 and as x approaches 2:
lbvjy [14]

Answer:

a = 4

b = -2

Step-by-step explanation:

If the given function is continuous at x = 1

\lim_{x \to 1^{-}} f(x)=(x+1)

                     =2

\lim_{x \to 1^{+}} f(x)=ax+b

                     =a+b

\lim_{x \to 1} f(x)=ax+b

                   =a+b

And for the continuity of the function at x = 1,

\lim_{x \to 1^{-}} f(x)=\lim_{x \to 1^{+}} f(x)=\lim_{x \to 1} f(x)

Therefore, (a + b) = 2 -------(1)

If the function 'f' is continuous at x = 2,

\lim_{x \to 2^{-}} f(x)=ax+b

                     =2a+b

\lim_{x \to 2^{+}} f(x)=3x

                     =6

\lim_{x \to 2} f(x)=3x

                   =6

Therefore, \lim_{x \to 2^{-}} f(x)=\lim_{x \to 2^{+}} f(x)=\lim_{x \to 2} f(x)

2a + b = 6 -----(2)

Subtract equation (1) from (2),

(2a + b) - (a + b) = 6 - 2

a = 4

From equation (1),

4 + b = 2

b = -2

3 0
3 years ago
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