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3 years ago

Help plzzzzzanswer correctly..

Mathematics

1 answer:

Sav [38]3 years ago

3 0

Answer:

I belive you need to calculate your heart rate first...

Step-by-step explanation:

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Can someone please help me with 1-11? Brainliest answer will be awarded! :)

vovangra [49]

The most important thing that you have to know is that all angles in a triangle equals 180°

8 0

3 years ago

If AB=48 and BC=22, then AC=?

alexandr1967 [171]

If A, B and C are collinear, then

1) if B is between A and C:

AC = AB + BC

AC = 48 + 22 = 70

2) if C is between A and B:

AB = AC + BC

48 = AC + 22 |-22

AC = 26

3) if A is between B and C:

BC = AB + AC

22 = 48 + AC |-48

AC = - 26 < 0 FALSE

Answer:

if B is between A and C, then AC = 70

if C is between A and B, then AC = 26

4 0

3 years ago

Read 2 more answers

Dan is comparing the cost of having printed and prepared by two companies

Kay [80]

For 4 copies, the total cost for preparing and printing will be the same.

Step-by-step explanation:

Let,

p be the number of pages

Company A;

Charges of preparing = $6

Charges of per page print = $2.50

A(p)= 2.50p + 6 Eqn 1

Company B;

Charges of preparing = $4

Charges of per page print = $3

B(p)= 3p + 4 Eqn 2

For equaling the cost;

A(p) = B(p)

$2.50p+6=3p+4\\2.50p-3p=4-6\\-0.50p=-2$

Dividing both sides by -0.50

$\frac{-0.50p}{-0.50}=\frac{-2}{-0.50}\\p=4$

For 4 copies, the total cost for preparing and printing will be the same.

Keywords: Addition, division

Learn more about division at:

#LearnwithBrainly

8 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2

Help plzzzzzanswer correctly..​

Help plzzzzzanswer correctly..