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3 years ago

Pls help will mark brainlest

Mathematics

1 answer:

kari74 [83]3 years ago

7 0

Answer:

15 bags

Step-by-step explanation:

She can fill 5 bags per each pound (because there are 5 $\frac{1}{5}ths$ per pound)

Since there are 3 pounds, and , again, each pound can fill 5 bags, she will need 15 bags.

Another way too look at it is to ask how many 1/5ths are in 3 pounds. To find this answer mathematically:

3 ÷ 1/5 = 15 because when we divide by a fraction, we need to flip it and then multiply (so 3 ÷ 1/5 is actually 3×5)

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The rate for one unit of a given quantity is called a(n)

ankoles [38]

The rate for one unit of a given quantity is called a(n) Unit Rate.

4 0

3 years ago

Please help I will give brainliest to whoever answers this question

pantera1 [17]

Hi!

To solve this, we must make an equation in y = mx + b form, where m is the slope and b is the y-intercept.

Since we are looking at a graph with points, it saves us a TON of work, and all we have to do is simply look at the graph. We only need to find m and b, that's it!

M is the slope, and the slope can be represented by the change between points on a line, also known as <em>rise/run. </em>

If we start at point (-4, -1), and go to point (-2, 0), we can see that it goes up 1 and right 2, which would be represented as 1/2.

B is the y-intercept, and that is simply the point that is on the y-axis, which is the point (0, 1), so our number would be 1.

Therefore, plugging these numbers into the equation, our equation is y = <em>1/2</em>x + <em>1</em>

Hope this helps! :D

6 0

2 years ago

PLEASE HELP WITH GEOMETRY IM STUCK!?}

Ilya [14]

Answer:

13.8

Step-by-step explanation:

Recall the trigonometric ratios

Sine = Opposite over Hypotenuse (SOH)

Cosine = Adjacent over Hypotenuse (CAH)

Tangent = Opposite over Adjacent (TOA)

Now lets look back at the question

We are given an angle and its adjacent side length (11) and we need to find the hypotenuse

The hypotenuse and adjacent sides corresponds with the trig function cosine so we will use cosine to solve for x

( remember that cosine = adjacent over hypotenuse )

$cos(37)=\frac{11}{x}$

step 1 multiply each side by x

$cos(37)*x=xcos(37)\\\frac{11}{x} *x=11$

now we have $xcos(37)=11$

step 2 divide each side by cos(37)

$\frac{xcos(37)}{cos(37)} =x\\\frac{11}{cos(37)} =13.77349\\$

we're left with x = 13.77439

Our last step would be to round to the nearest tenth

We would get that the answer is 13.8

3 0

3 years ago

Read 2 more answers

What is the y-coordinate of the midpoint of the line segment of (-4,13) and (18,-3.5)

Nady [450]

The midpoint is (7,4.75) so the y-coordinate should be y=4.75. I hope that helps

4 0

3 years ago

In a certain region of space, the potential is given by v=2x−5x2y 3yz2. How much is the magnitude of the electric field at point

Masja [62]

The magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

<h3>How to calculate the electric field?</h3>

Since in a certain region of space, the potential is given by v = 2x − 5x²y 3yz²,

The electric field, E = -grad(V) where grad(V) = (dV/dx)i + (dV/dy)j + (dV/dz)k

Since V = 2x − 5x²y + 3yz²

dV/dx = d(2x − 5x²y + 3yz²)/dx

= d2x/dx - d5x²y/dx + d3yz²/dx

= 2 - 10xy + 0

= 2 - 10xy

dV/dy = d(2x − 5x²y + 3yz²)/dy

= d2x/dy - d5x²y/dy + d3yz²/dy

= 0 - 5x² + 3z²

= - 5x² + 3z²

dV/dz = d(2x − 5x²y + 3yz²)/dz

= d2x/dz - d5x²y/dz + d3yz²/dz

= 0 - 0 + 6z

= 6z

<h3>The electric field</h3>

So, E = -grad(V)

= -[(dV/dx)i + (dV/dy)j + (dV/dz)k]

= -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

So, the electric field at (-2, 2, 0) is

E = -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

E = -[(2 - 10(-2)(2))i + (- 5(2)² + 3(0)²)j + (6(0))k]

E = -[(2 + 40)i + (- 5(4) + 0)j + (0)k]

E = -[42i + (-20 + 0)j + (0)k]

E = -[42i - 20j + 0k]

E = -42i + 20j - 0k

<h3>The magnitude of the electric field</h3>

So, the magnitude of E at (-2, 2, 0) is

|E| = √[(-42)² + 20² + 0²]

= √[1764 + 400 + 0]

= √2164

= 46.52 V/m

So, the magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

Learn more about electric field here:

brainly.com/question/25751825

#SPJ12

3 0

2 years ago