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Rainbow [258]
3 years ago
6

Pls help! ASAP!

Mathematics
2 answers:
Anastasy [175]3 years ago
4 0

Answer:

I think the answer would be high.

Step-by-step explanation:

You can see how the X on the bottom is (13) it's one less than the X on the top. So my reasoning is since there are fewer differences b/w the two sets, there are more vaiables overlapping.

seraphim [82]3 years ago
4 0
It’s high girlie!!! <3333333333333
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I need help please and thank you
dem82 [27]
So we see that the ratio between the two octagons is 7:1, since 28/4=7 so what we do next is multiply the values of the smaller octagon by 7. But that’s the long way. There’s actually a shortcut by multiplying the perimeter of the smaller octagon, 34, by 7. This in turn equals 238.
6 0
3 years ago
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Paige has 4 more balloons than Zoee. Emma  has 2 more balloons than Paige. Altogether they have 100 balloons. How many does each
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3 years ago
A wheel with radius 1 foot makes 1 revolution in 4 seconds. What is the linear velocity, in feet per second, of a point on the e
Alona [7]

In one revolution of the wheel, a point on the edge travels a distance equal to the circumference of the wheel.

The wheel has radius 1 ft, so its circumference is 2π (1 ft) = 2π ft.

Then the point has a linear speed of

(1/4 rev/s) * (2π ft/rev) = 2π/4 ft/s = π/2 ft/s

5 0
3 years ago
Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators
Nata [24]

Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

where p(x) and h(y) are continuous functions.

We have the following differential equation

y'=(1-5x)y^2, \quad y(0)=-12

In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

4 0
3 years ago
Assume the triangular prism has a base area of 49 cm^2 and a volume of 588 cm^3. Which base side length does the rectangular pri
andre [41]

Answer:

Length = Width = 7 cm

Step-by-step explanation:

8 0
3 years ago
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