2 years ago

What’s the answer of this problem

Mathematics

1 answer:

Elden [556K]2 years ago

3 0

Answer:

no b 11 sq.units is the answer

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-<br><img src="https://tex.z-dn.net/?f=%20-%203%5Cpi%20%2B%20w%20%3D%202%5Cpi" id="TexFormula1" title=" - 3\pi + w = 2\pi" alt="

pantera1 [17]

Isolate the w. Note the equal sign. What you do to one side, you do to the other.

Add 3π to both sides

- 3π (+3π) + w = 2π (+3π)

w = 2π + (3π)

w = 5π

w = 5π is your answer

hope this helps

4 0

3 years ago

A DISASTER RELIEF GROUP DELIVERS 395 FIRST AID KITS. THE GROUP DELIVERS AN EQUAL NUMBER OF FIRST AID KITS TO 3 COUNTRIES. HOW MA

postnew [5]

395/3
about 132
Hope this helped

7 0

3 years ago

Read 2 more answers

A proportional relationship is graphed on the coordinate plane. The line that represents the relationship passes through points

Stels [109]

Answer:

Step-by-step explanation:

HELLPPPP

4 0

3 years ago

Solve for ppp. 16-3p=\dfrac23p+516−3p= 3 2 p+516, minus, 3, p, equals, start fraction, 2, divided by, 3, end fraction, p, plus

Ksivusya [100]

Given:

The given equation is:

$16-3p=\dfrac{2}{3}p+5$

To find:

The value of p.

Solution:

We have,

$16-3p=\dfrac{2}{3}p+5$

Multiply both sides by 3.

$3(16-3p)=3\left(\dfrac{2}{3}p+5\right)$

$48-9p=2p+15$

Isolating the variable terms, we get

$48-15=2p+9p$

$33=11p$

Divide both sides by 11, we get

$\dfrac{33}{11}=p$

$3=p$

Therefore, the required solution is $p=3$ .

7 0

3 years ago

Find an equation for the plane that is tangent to the surface z equals ln (x plus y )at the point Upper P (1 comma 0 comma 0 ).

alexira [117]

Let $f(x,y,z)=z-\ln(x+y)$ . The gradient of $f$ at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.

So the tangent plane has equation

$\nabla f(1,0,0)\cdot(x-1,y,z)=0$

Compute the gradient:

$\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)$

Evaluate the gradient at the given point:

$\nabla f(1,0,0)=(-1,-1,1)$

Then the equation of the tangent plane is

$(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}$

7 0

4 years ago