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iragen [17]
3 years ago
6

which shape had a larger area: a rectangle that is 7 inches by 3/4 inch, or a square with side length of 2.5 inches? Show your r

easoning.
Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0
In order to find the area for a rectangle, you must multiply length times width, which would be 7 x 3/4 which is 5.25. In order to find the area for a square, you do length^2, also know as length x length. Which would be 2.5^2 which is 2.5 x 2.5 which is 6.25. 6.25 is larger than 5.25, so the square has a larger area.
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Factor the following , it is not prime
MrMuchimi
So 1/9-16z^2 this is a diffirence of two perfect squares thing
so (1/3)^2-(4z)^2
so ((1/3)-4z)((1/3)+4z)

4 0
3 years ago
V is greater than or equal to 5
Citrus2011 [14]

Answer:

greater

Step-by-step explanation:

3 0
3 years ago
A gardener already has four and one over 2 ft of fencing in his garage. He wants to fence in a square garden for his flowers. Th
NikAS [45]
<h2>Answer:</h2>

6\frac{1}{2}<u>, "six and one over two ft".</u>

<h2>Step-by-step explanation:</h2>

Let's considerate the fact that the garden has a <u>square shape</u>.

<h3>1. Finding values of interest.</h3>

Amount of fence that the gardener already has: 4\frac{1}{2} ft.

Length of one side: 2\frac{3}{4} ft.

If one side measures 2\frac{3}{4} ft, and the square garden has 4 sides of equal length, because it's a square, then we must multiply the measure of one side by 4 to find the total length of fence needed:

4*(2\frac{3}{4})=\\ \\4*(2+\frac{3}{4})=\\ \\(4*2)+(4*\frac{3}{4})=\\ \\8+(\frac{12}{4} )=\\ \\8+3=\\ \\11

<h3>2. How much more does he need?</h3>

The gardener already has  4\frac{1}{2} , which equals 4 + \frac{1}{2}. Hence, the difference between the amount needed and the amount that the gardeneralready has will give us the remaining amount required. Let's do that:

11-(4+\frac{1}{2} )=\\ \\11-(\frac{8}{2} +\frac{1}{2} )=\\ \\11-\frac{9}{2}= \\ \\\frac{22}{2} -\frac{9}{2}=\\\\ \frac{13}{2}

<h3>3. Express your result.</h3>

\frac{13}{2} =\\ \\\frac{2}{2} +\frac{2}{2} +\frac{2}{2} +\frac{2}{2} +\frac{2}{2} +\frac{2}{2} +\frac{1}{2}= \\ \\6+\frac{1}{2}=\\ \\6\frac{1}{2}

8 0
2 years ago
If the point P(-4/5,y) lies on the unit circle and P is in the third quadrant, what does y equal? If necessary, use the slash ma
aleksandr82 [10.1K]
We use the fact that x2+y2=1  This is the equation of the unit circle.
We know that x=-4/5, so plug this in and solve for y
x2+y2=1 <span>(−4/5)^2+y2=1</span> 16/25+y2=1 y2=1−16/25 y2=9/25 <span>y=−square root of 9/ 25</span>
y = -3/5
8 0
3 years ago
Read 2 more answers
Solve the given differential equation by using an appropriate substitution. the de is homogeneous. (y2 yx) dx − x2 dy = 0
Schach [20]

The solution for the given differential equation is lnlxl = \frac{-x}{y} +C

Given,

(y^{2} +y^{x} )dx - x^{2} dy =0

Here,

x = vy, dx = vdy + ydv

y = ux, dy = udx + xdu

Then,

((ux)^{2} +(ux)x)dx-x^{2} (udx+xdu)=0\\=u^{2} x^{2} dx+ux^{2} dx-x^{2} udx-x^{3} du=0\\=u^{2} x^{2} dx=x^{3} du\\=\frac{x^{2} }{x^{3} } dx=\frac{1}{u^{2} } du

Now,

\int\limits^a_b {\frac{1}{x} } \, dx =\int\limits^a_b {u^{-2} } \, du

lnlxl=\frac{u^{-1} }{(-1)} +C

lnlxl=\frac{-1}{u} +C

y = ux

u = \frac{y}{x}

That is lnlxl =\frac{-x}{y} +C

Learn more about differential equation here: brainly.com/question/21852102

#SPJ4

8 0
1 year ago
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