Which of the following is a perfect square? A. 127 B. 100 C. 102 D. 13

Answers

Answer 1
Answer:

Hi there


The answer is B , 100

because

√100= 10 (That is a perfect square)


I hope that's help !


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If f(x) = 2(x)^2 + 5 square root (x+2), complete the following statement f(2) = ______

Answers

f(x) is a fancy term for saying y. It is mainly used in Trigonometry, Pre-Calculus, Calculus, and beyond.

f(x) = 2x^2 + 5√(x+2)

plug in the 2 into the x.

f(2)= 2(2)^2 + 5√(2+2)

this gives:

f(2) = 8 + 5√(4)

square the 4.

f(2) = 8 + 5(2)

multiply 5 and 2.

f(2) = 8 + 10

add them together.

f(2) = 18

Glad to help. 


Final answer:

By substituting x = 2 into the function f(x) = 2(x)² + 5√(x + 2) and simplifying the equation, the resulting value of f(2) is calculated to be 18.

Explanation:

For the given function f(x) = 2(x)² + 5√(x + 2), we are asked to find the value of f(2). To find this, first substitute x = 2 into the function. So, the calculation becomes

f(2) = 2 * (2)² + 5 * √(2 + 2)

simplifying this equation, we get:

f(2) = 2 * 4 + 5 * √4 = 8 + 5 * 2 = 8 + 10 = 18

So, f(2) = 18

Learn more about Function evaluation here:

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The number of dogs in a shelter is equal to 4 less than 2 times the number of cats. If there are 20 dogs in the shelter how many cats are there?

Answers

d=number of dogs
c=cats

d equals 4 less than 2 times c
d=-4+2c
20 dogs
d=20
subsitute
20=-4+2c
add 4
24=2c
divide 2
12=c
12 cats

Tahmar knows the formula for simple interest is I = Prt, where I represents the simple interest on an amount of money, P, for t years at r rate. She transforms the equation to isolate P : P = . Using this formula, what is the amount of money, P, that will generate $20 at a 5% interest rate over 5 years?

Answers

 $80

Working
Making p the subject gives:
P=
(I)/(rT)
Substituting for values of I, r and t
p= (20)/(0.05*5)
p=80 dollars

Answer: 80

Step-by-step explanation: The given formula is: P = I/rt. So, we'll plug in $20, 5%, and 5 years like this: P = 20/5%(5) (or for the 5%, you can put .05), you will get: 80. You can get that answer on your calculator as well.

Hope this helped :)

How do you solve 5[2p−4(p+5)]=25 properly?

Answers

Answer:

Step-by-step explanation:

5[2p-4(p+5)]=25

or, 2p-4(p+5)=5

or, 2p-4p-20=5

or, -2p=5+20

or, -2p=25

or, 2p=-25

or, p=25/2

so, p=12.5

Solve this linear equations: x + y + z = 34 1x + 10y + 5z = 100

Answers

Answer

To solve this system of linear equations, we can use the method of substitution.

First, let's solve the first equation for x:

x = 34 - y - z

Now, we substitute this value of x into the second equation:

1(34 - y - z) + 10y + 5z = 100

34 - y - z + 10y + 5z = 100

34 + 9y + 4z = 100

Next, we simplify the second equation:

9y + 4z = 100 - 34

9y + 4z = 66

We can rewrite this equation as:

9y = 66 - 4z

y = (66 - 4z) / 9

Now, we substitute this value of y back into the first equation:

x + (66 - 4z) / 9 + z = 34

Multiplying through by 9 to eliminate the fraction:

9x + 66 - 4z + 9z = 306

9x + 5z = 240

Now we have a system of two equations in two variables:

9x + 5z = 240

9y + 4z = 66

We can solve this using the method of substitution or elimination. Let's use the method of elimination:

Multiplying the first equation by 4 and the second equation by 5, we get:

36x + 20z = 960

45y + 20z = 330

Subtracting the second equation from the first, we eliminate z:

36x - 45y = 630

We can simplify this equation by dividing through by 9:

4x - 5y = 70

Now, let's solve the new system of equations:

4x - 5y = 70

9y + 4z = 66

We can multiply the first equation by 9 and the second equation by 4 to eliminate x:

36x - 45y = 630

36y + 16z = 264

Now, subtracting the first equation from the second, we eliminate y:

36y + 16z - 36x + 45y = 264 - 630

81y + 16z = -366

Dividing through by 3, we get:

27y + 16z = -122

Now, we have a system of two equations in two variables:

4x - 5y = 70

27y + 16z = -122

We can solve this system using the method of substitution or elimination.

What is 3ft to 7 yd as a fraction in simplest form

Answers

3/7 or three sevenths