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3 years ago

Which is it? please help me and i will give brainliest

Mathematics

2 answers:

Allisa [31]3 years ago

4 0

Answer:

it is 1/2 i.e. the first option

this is simple ratio method;

white : brown

2 : 1

1 : x

x = 1/2 cups

Hope this helps!

I would really appreciate if you mark me brainliest as I have been trying to get there for a veryyy long time now :)

Cerrena [4.2K]3 years ago

3 0

Answer: A, 1/2

Step-by-step explanation:

he is simply using 1 cup of bs for each cup of ws so when you do it its i cup of ws then its hald of 1 bs

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How does the number of decimal places in the factors relate to the number of decimal places in the product of the example

Nostrana [21]

To multiply decimals: Set up and multiply the numbers as you do with whole numbers. Count the total number of decimal places in both of the factors. Place the decimal point in the product so that the number of decimal places in the product is the sum of the decimal places in the factors.

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3 years ago

-9+n/4=-15 answer????

victus00 [196]

Answer:

-2

Step-by-step explanation:

9+n/4= -15

cross multiply

9+4-15=n

13-15=n

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2 years ago

Can someone help me find the relative maximum and minimum? please i really need help

bearhunter [10]

9514 1404 393

Answer:

relative maximum: -4
relative (and absolute) minimum: -5

Step-by-step explanation:

The curve has a relative maximum where values on either side are lower. This looks like a peak in the curve. There is one of those on the y-axis at y = -4.

The relative maximum is -4.

A relative minimum is a low point, where the curve is higher on either side. There are two of these, located symmetrically about the y-axis. The minimum appears to be about y = -5. (They might be at x = ± 1, but it is hard to tell.)

The relative minima are -5.

A minimum or maximum is absolute if no part of the curve is lower or higher. Here, the minima are absolute, while the maximum is only relative. (The left and right branches of the curve go higher than y=-4.)

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Identifying the points on the curve should be the easy part. Deciding what the coordinates are can be harder when the graph is like this one.

4 0

2 years ago

Question in the picture

ehidna [41]

Answer:

y=2x+5 is the answer ALSO CAN WE TALK ABOUT THE FACT THAT MOO975 IS ON THREE QUESTIONS AT THE SAME TIME

Step-by-step explanation:

6 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago