MATHEMATICS HIGH SCHOOL

Sam determines the zeros of the function f(x) to be 8 and 7. What could be Sam’s function?1.

f(x) = (x − 8)(x + 7)
2.

f(x) = (x − 8)(x − 7)
3.

f(x) = (x + 8)(x + 7)
4.

f(x) = (x + 8)(x − 7)

Answers

Answer 1
Answer:

Answer:

2. f (x) = (x - 8)\cdot (x-7)

Step-by-step explanation:

Given that function is polynomial, each zero is contained in binomials of the form (x - a), where a is the value of the zero. The number of roots means the number of binomial and the grade of the polynomial. Then, the function has the following form:

f (x) = (x - 8)\cdot (x-7)

In cosequence, the right answer is 2.
Answer 2
Answer: Option 2: f(x) = (x-8)(x-7)

x-8 = 0 ⇒ x = 8

x-7 = 0 ⇒ x = 7

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The product of 2.5 and a number t

Answers

The word 'product' tells you that your two numbers will be multiply. When you multiply a number and a variable, the number always goes first. So, the product of 2.5 and a number t is 2.5t . 

Help me please !!!!!!

Answers

If you read the Question carefully again, you will see that they mention about fence all around the garden (perimeter) which is 256ft. Therefore, you can create an equation for it:
(W+8)+W=256
2W+8 = 256
2W= 256-8
2W = 248
W= 248/2
W= 124

Adding which terms to 3x2y would result in a monomial? Check all that apply.3xy


–12x2y


2x2y2


7xy2


–10x2


4x2y


3x3

Answers

Adding and subtracting amonomial requires having the same variables. No matter how big or small theircoefficient is, if their variables do not match, they cannot be added or subtracted.The crucial part in adding or subtracting monomials is their sign. If the signsare the same, retain the sign. If the signs are different, subtract and keepthe sign of the larger number.

3x2y + 3xy (cannot be added)
3x2y + (–12x2y) = -9x2y  
3x2y + 2x2y2 (cannot be added)
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Answer:

B and F would be your correct answers

Which numbers are factors of 40? select all that apply. 4 10 7 80 submit?

Answers

The factors of 40 from that list are 4, 8. and 80. A factor needs to be able to be divided by 40.

A colony contains 1500 bacteria. The population increases at a rate of 115% each hour. If x represents the number of hours elapsed, which function represents the scenario?A f(x) = 1500(1.15)x

B f(x) = 1500(115)x

C f(x) = 1500(2.15)x

D f(x) = 1500(215)x

Answers

Answer:

Option (c) is correct.

f(x)=1500(2.15)^x function representing the increase of bacteria every hour x,

Step-by-step explanation:

Given : A colony contains 1500 bacteria. The population increases at a rate of 115% each hour.

we have to find the function that represents the given scenario.

Let x represents the number of hours elapsed.

Given  A colony contains 1500 bacteria

 and number of bacteria is increasing at a rate of 115% each hour.

Using formula for Compound interest , we have,

A=P(1+(r)/(100) )^t

Where A is amount

T is time period

R is rate of interest

Here, P = 1500

T = x hours

R = 115%

Let f(x) be the function representing the increase of bacteria every hour.

Substitute, we have,

f(x)=1500(1+(115)/(100) )^x

Simplify, we get,

f(x)=1500(1+1.15)^x

f(x)=1500(2.15)^x

Thus, f(x)=1500(2.15)^x function representing the increase of bacteria every hour x,
its  f(x)=1500(2.15)x so c hope this helps

Divide 7/15 by 3/5.
A. 75/21
B. 7/9
C. 7/25
D. 21/75

Answers

In this question there is nothing complicated. Only thing is to know the way fractions can be divided. Once that is known the problem would be one of the easiest to solve. Now let us get back to the problem and look at all the information's that are given in the question.
Divide 7/15 by 3/5 = (7/15)/(3/5)
                              = (7 * 5)/(15 * 3)
                               = (35/45)
Dividing the numerator and the denominator by 5 for simplifying purpose, we get
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So from the above deduction we can easily conclude that 7/9 is the correct answer and option "B" is the correct option among all the options given in the question.
If you would like to solve 7/15 / 3/5, you can calculate this using the following steps:

7/15 / 3/5 = 7/15 * 5/3 = 7/9

The correct result would be B. 7/9.
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