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Roman55 [17]
3 years ago
5

What is the value of the expression 2^2+6^2÷2^2​

Mathematics
1 answer:
Studentka2010 [4]3 years ago
3 0

Answer:

13

Step-by-step explanation:

2^{2} + 6^{2} / 2^{2} = 4+ 36/4 = 4 +9 = 13

0r

4 + (6/2)^2 = 4+ 3^2 = 4+9 = 13

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Sadie is 3 years older than Kareem, and Kareem is twice as old as Nadra. The sum of their ages is 33.
Dima020 [189]

Answer:

Step 1-Write an equation, letting x represent Kareem’s age:

(x + 3) + x + 2 x = 33

Step 2-Solve the equation:

(x + 3) + x + 2 x = 33. 3 + 4 x = 33. 4 x = 30. x = 7 1/2.

Step 3-Sadie is 10 1/2, Kareem is 7 1/2, and Nadra is 3 3/4.

Which corrects the student’s first error?

1. In Step 1, the student should have used One-half x to represent Nadra’s age, and written the equation (x + 3) + x +1/2x = 33.

2. In Step 2, the student should have added 3 on both sides of the equation to find a solution of x = 9.

3. In Step 3, the solution represents Sadie’s age. Therefore, Sadie is 7 1/2, Kareem is 4 1/2, and Nadra is 2 1/4.

4. In Step 3, Nadra’s age is the difference between 33 and Sadie’s and Kareem’s ages, so Nadra is 15.

5 0
2 years ago
Select two ratios that are equivalent to 27: 9.<br> Choose 2 answers:
schepotkina [342]

Answer:

1. 9:3

2. 3:1

Step-by-step explanation:

Hope it really helps you

5 0
3 years ago
What is the rise over run​
Dovator [93]

Rise over run is another term for slope, with we can use to derive a linear equation.

The term rise over run in technical terms is the change of y over the change of x.

To find the rise over run, subtract the y terms and divide that by the x terms.

\frac{y_2-y_1}{x_2-x_1}

You can use this to derive a linear equation as earlier mentioned by plugging in given points.

For example, a line passes through (3,4) with a rise over run of 3.

y=mx+b

4=3(3)+b

4=9+b

b=-5

So therefore the y intercept is -5, and the equation is y=3x-5

5 0
3 years ago
I need to verify identity functions for a one to one function.
Eddi Din [679]

We have the function

f(x)=(x+6)^3

1. For f^-1:

Let y = f(x) = (x+6)^3

Switch x and y to get:

x=(y+6)^3

And solve for y

\begin{gathered} x^{\frac{1}{3}}=y+6 \\ x^{\frac{1}{3}}-6=y+6-6 \\ x^{\frac{1}{3}}-6=y \end{gathered}

And we have y = f^-1(x)

Answer blank 1:

f^{-1}(x)=x^{\frac{1}{3}}-6

2. For f o f^-1 (x):

(f\circ f^{-1})(x)=f(f^{-1}(x))

And solve

\begin{gathered} =f(x^{\frac{1}{3}}-6) \\ =(x^{\frac{1}{3}}-6+6)^3 \\ =(x^{\frac{1}{3}})^3 \\ =x \end{gathered}

answer blank 2

x^{\frac{1}{3}}-6

answer blank 3

x^{\frac{1}{3}}-6

answer blank 4

x^{\frac{1}{3}}

3. For f^-1 o f:

(f^{-1}\circ f)(x)=f^{-1}(f(x))

Solve

\begin{gathered} =f^{-1}((x+6)^3) \\ =\sqrt[3]{(x+6)^3}-6 \\ =x+6-6 \\ =x \end{gathered}

answer blank 5

(x+6)^3

answer blank 6

(x+6)^3

answer blank 7

x+6

4 0
1 year ago
Can someone help me please
JulsSmile [24]

Answer:

yes

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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