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3 years ago

It also detects if you are right or wrong

Mathematics

2 answers:

harkovskaia [24]3 years ago

6 0

It’s the letter A because he spends money, not get money, so C is def not right.

And B just doesn’t make sense.

Sindrei [870]3 years ago

3 0

Answer:

I think A

Step-by-step explanation:

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Find the value of c that makes the expression a perfect square trinomial.<br><br> x2 + 4x + c

masha68 [24]

the value of c that makes the expression a perfect square binomial is c=4 .

<u>Step-by-step explanation:</u>

Here we have , an expression x2 + 4x + c or , $x^2 + 4x + c$ . We need to find the value of c that makes the expression a perfect square binomial. Let's find out:

We have , $x^2 + 4x + c$

⇒ $x^2+4x+c$

⇒ $x^2+2(2)x+c$

Now , we know that $(a+b)^2=a^2+2(a)(b)+b^2$

Comparing above equation , to $x^2+2(2)x+c$ we get ;

⇒ $x^2+2(2)x+c$ { $c=2^2$ }

⇒ $x^2+2(2)x+2^2$

⇒ $(x+2)^2$

Therefore , the value of c that makes the expression a perfect square binomial is c=4 .

7 0

3 years ago

Alicia and Bennie have decided to play a game. They each draw as many flowers on paper as they can within one minute, repeating

ziro4ka [17]

Answer:

Because The Flowers Bennie drew are independent

3 0

2 years ago

Read 2 more answers

Mary Hernandez had a policy with a $500 deductible which paid 80% of her covered charges less deductible. She had medical expens

12345 [234]

A.) The insurance company's payment is 80% of the total minus the $500 deductible, so:
(10000 - 500) * .8 = 7600

B.) To find the Copay, you do almost the same thing:
(10000 - 500) * .2 = 1900

C.) Mary's Total Cost: 1900 + 500 = 2400.00

4 0

3 years ago

How to solve 5/2 x X=1

IgorLugansk [536]

2.5 is the answer u welcome

7 0

3 years ago

Suppose that we roll a red and a black die. Let a = "the black die shows a 2 or a 5", b = "the sum of the two dice is at least 7

Novosadov [1.4K]

Answer: No, A and B are not independent events.

∵ it does not satisfy the rule of probability for independent events i.e.

P(A∩B)=P(A).P(B)

Explanation:

Let A be the event that the black dice shows 2 or 5

Let B be the event that the sum of two dice is atleast 7

Sample space of A={ $(R_1,B_2)(R_2,B_2)(R_3,B_2)(R_4,B_2)$ ( $(R_5,B_2),(R_6,B_2),(R_1,B_5),(R_2,B_5),(R_3,B_5),(R_4,B_5),(R_5,B_5),(R_6,B_5)$ }

Sample space of B= { $(R_1,B_6),(R_2,B_5),(R_2,B_6),(R_3,B_4),(R_3,B_5),(R_3,B_6),$ , $(R_4,B_3),(R_4,B_4),(R_4,B_5),(R_4,B_6),(R_5,B_2),(R_5,B_3),(R_5,B_4),(R_5,B_5)$ , $(R_5,B_6),(R_6,B_2),(R_6,B_3),(R_6,B_4),(R_6,B_5),(R_6,B_5)$ }