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Romashka-Z-Leto [24]

2 years ago

15

ms. beck divided 2 cups of sugar equally into 8 containers. what fraction of a cup of sugar did each container receive?

2 answers:

Bond [772]2 years ago

5 0

1/4 cup in each container. divide 2 by 8 and you get .25 or 25% or 1/4

ExtremeBDS [4]2 years ago

4 0

Solving this out you would do 8 divides by 2 and it would give you 4

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What is (-8) x (-2/3)

Triss [41]

Answer:

16/3 or 5.3 recurring

Step-by-step explanation:

-8 times -2/3 is 16/3, and 16/3 is 5.3 recurring

7 0

2 years ago

Is this correct question correct?

____ [38]

No, it is not correct.

First of all, you put the little 4 after the 8.

What you did in the equation is 8x4, not 8x8x8x8.

So, the correct answer to 8x8x8x8 would be 4096.

<em><u>Any Questions? Comment Below!</u></em>

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4 0

3 years ago

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Dan weighs 205 pounds but is only 5 feet 8 inches tall. Evan is 6 feet tall. How much would you expect Evan to weigh if they hav

Rina8888 [55]

Answer:

Around 217 pounds

Step-by-step explanation:

Let's convert the height into inches.

5 feet 8 = $5\cdot12 + 8 = 60 + 8 = 68$

6 feet $= 6\cdot12 = 72$ .

We can set up a proportion

$\frac{205}{68} = \frac{x}{72}$

We can use the cross products property to find x.

$205\cdot72=14760\\\\\\14760\div68\approx217$

Hope this helped!

4 0

2 years ago

Read 2 more answers

Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

so that the left side condenses into the derivative of a product.

$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

Integrate both sides, then solve for $v$ , then for $y$ :

$x^3v=\displaystyle\int3x^2\,\mathrm dx$

$x^3v=x^3+C$

$v=1+\dfrac C{x^3}$

$y^3=1+\dfrac C{x^3}$

$\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}$

6 0

3 years ago

Can someone help me with this question?

KonstantinChe [14]

-3. Put the two equation equal to each other and solve.
4x+18=2x+12
-2x=-2x
2x+18=12
-18=-18
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x=-3

8 0

2 years ago