All categories

3 years ago

PLEASE ANSWER THIS FAST!!!!!!

Mathematics

2 answers:

o-na [289]3 years ago

7 0

Answer:

Step-by-step explanation:

The correct answer is B

GK = 5 and KF = 5

5/5 = 1

HJ = 3

JG = 3

HJ/JG = 1

OLga [1]3 years ago

6 0

Answer:

Step-by-step explanation:

It appears that only B is the correct option, since $\frac{5}{5} =\frac{3}{3} = 1$

You might be interested in

BRAINLY!! 20 POINTS!!

mario62 [17]

A. the mean remains the same.

For more info: brainly.com/question/4249234

Hope this helps love!

7 0

3 years ago

Read 2 more answers

Kelvin saves $60 a day. Al saves $90 a day. How many times Kelvin's savings are Al's savings?

vaieri [72.5K]

How many times Kevin's savings are al's savings is 30 more

5 0

3 years ago

Read 2 more answers

Write an equivalent algebraic equation for the sentence below.

tatuchka [14]

Answer:

x-7=13

Step-by-step explanation:

x-7= 13

hope that helps

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=2%20-%206%20%5Cdiv%206" id="TexFormula1" title="2 - 6 \div 6" alt="2 - 6 \div 6" align="absmid

Karolina [17]

2 - 6 ÷ 6 (Order of Operations - PEMDAS)

= 2 - 1

= 1

Answer is 1

7 0

3 years ago

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$

4 0

3 years ago