All categories

2 years ago

Pls help thannnnkkkss <33

Mathematics

1 answer:

juin [17]2 years ago

5 0

Answer: You would put the parentheses here 24 divided by 6 x 2 - (1 +4)

Have a nice day!

You might be interested in

What is the slope? First, determine the slope. (200,14) (225,16)

Maksim231197 [3]

Answer:

$\displaystyle m=\frac{2}{25}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Slope Formula: $\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (200, 14)

Point (225, 16)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m

Substitute in points [SF]: $\displaystyle m=\frac{16-14}{225-200}$
[Fraction] Subtract: $\displaystyle m=\frac{2}{25}$

7 0

2 years ago

Hiiii could one of y’all help please. be 100% correct i’m having a hard time

ivanzaharov [21]

Answer:

other

Step-by-step explanation:

4 0

2 years ago

Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =

dexar [7]

Given $f(x,\ y)=x^2-y^2-2$ subject to the constraint $x^2+y^2=16$

Let $g(x,\ y)=x^2+y^2$ .

The gradient vectors of $f$ and $g$ are:

$\nabla f(x,\ y)=\langle2x,-2y\rangle$ and $\nabla g(x,\ y)=\langle2x,2y\rangle$

By Lagrange's theorem, there is a number $\lambda$ , such that

$\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle$

$\lambda=\pm1$

It can be seen that $f(x,\ y)=x^2-y^2-2$ has local extreme values at the given region.

6 0

2 years ago

Estimate the sum : 7.2 + 7.3 + 6.9 + 6.8.

kkurt [141]

Answer:

28.2

Step-by-step explanation:

7.2 + 7.3 + 6.9 + 6.8 = 28.2

8 0

2 years ago

Read 2 more answers

What conditions would be enough to prove that P is the circumcenter of HJK ? Select all that apply.

nadezda [96]

Answer:

B trust me

Step-by-step explanation:

6 0

1 year ago