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3 years ago

Please help me fast plz

Mathematics

1 answer:

mafiozo [28]3 years ago

4 0

Answer:

First Bag: 15kg

Second Bag: 22.5kg

Third Bag: 52.5kg

Step-by-step explanation:

90 / 12 = 7.5

2 x 7.5 = 15

3 x 7.5 = 22.5

7 x 7.5 = 52.5

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Answer:

11 degrees, i think?

Step-by-step explanation:

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3 years ago

If possible please help! With 12

madreJ [45]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

F=(2xy +z³)i + x³j + 3xz²k find a scalar potential and work done in moving an object in the field from (1,-2,1) to (3,1,4)

Alex73 [517]

Step-by-step explanation:

Given:

$\textbf{F} = (2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}$

This field will have a scalar potential $\varphi$ if it satisfies the condition $\nabla \times \textbf{F}=0$ . While the first x- and y- components of $\nabla \times \textbf{F}$ are satisfied, the z-component doesn't.

$(\nabla \times \textbf{F})_z = \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right)$

$\:\:\:\:\:\:\:\:\: = 3x^2 - 2x \ne 0$

Therefore the field is nonconservative so it has no scalar potential. We can still calculate the work done by defining the position vector $\vec{\textbf{r}}$ as

$\vec{\textbf{r}} = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}$

and its differential is

$\textbf{d} \vec{\textbf{r}} = dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}}$

The work done then is given by

$\displaystyle \oint_c \vec{\textbf{F}} • \textbf{d} \vec{\textbf{r}} = \int ((2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}) • (dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}})$

$\displaystyle = (x^2y + xz^3) + x^3y + xz^3|_{(1, -2, 1)}^{(3, 1, 4)}$

$= 422$

5 0

3 years ago

Can someone please explain how to do this

Nookie1986 [14]

Answer:

the mid point formula for this is x+x/2 and y+y/ 2 so #9would be 1/2, 3/2

7 0

3 years ago

Help please! show work thanks

Andre45 [30]

Answers
b = 2.77 m
A = 43.0°
C = 111.1°

cosine law to find b

$b^2 = a^2 + c^2 -2ac \cos B \\ \\
b = \sqrt{a^2 + c^2 -2ac \cos B} \\ \\
b = \sqrt{4.33^2 + 5.92^2 - 2(4.33)(5.92) \cos 25.9} \\
b = 2.7708\ m$

b = 2.7708\ m

Find angle A with sine law

$\displaystyle
\frac{\sin A}{a} = \frac{\sin B}{b} \\ \\
\sin A = \frac{a \sin B}{b} \\ \\
A = \sin^{-1} \left[ \frac{a \sin B}{b} \right] \\ \\
A = \sin^{-1} \left[ \frac{4.33 \sin 25.9}{2.7708} \right] \\ \\
A = 43.0467020$

Find C with angles in triangle sum to 180

A + B + C = 180
C = 180 - A - B
C = 180 - 43.0467020 - 25.9
C = 111.1

6 0

4 years ago