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drek231 [11]
3 years ago
5

Ribbon A is 1/3 meters long. It is 2/5 meters shorter than ribbon b. What's the total length of 2 ribbons?

Mathematics
1 answer:
Pavel [41]3 years ago
4 0
1/3+2/5=5/15+10/15
answer is 1
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The diagram below shows the rectangle PLUM.
Illusion [34]

Answer:

Area of rectangle PLUM = 75.00 square units

Step-by-step explanation:

Since diagonal of rectangle divides the rectangle into two equal triangles,

Therefore, area of the rectangle PLUM = 2× area of triangle PLM

By the mean proportional theorem,

In ΔPLM,

AP² = AM × AL

6² = AM × 8

AM = \frac{36}{8}

AM = 4.5 units

Area of PLM = \frac{1}{2}(\text{Base})(\text{height})

                     = \frac{1}{2}(ML)(AP)

                     = \frac{1}{2}(8 + 4.5)(6)

                     = 12.5 × 3

                     = 37.5 units²

Now area of rectangle PLUM = 2×37.5 = 75 units²

Therefore, area of the rectangle is 75.00 square units.

3 0
3 years ago
In △ABC point D is the midpoint of AB , point E is the midpoint of BC , and point F is the midpoint of BE . Find area of △ABC, i
xz_007 [3.2K]

I did this   manually on  some notepaper but my scanner is not working.

The answer is 72 cm^2

3 0
3 years ago
Gcf using prime factorization of 52 and 65
Leto [7]
<span>25: 2 × 2 × 13 65: 5 × 13</span>
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3 years ago
Andrew claims the initial value and y - intercept are the same thing on a graph. Is he correct?
Agata [3.3K]

Answer:

We conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

Step-by-step explanation:

We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.

In other words, the initial value is the y-value or output of the point at x = 0

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the initial value of the equation y = 2x+1 is: y = 1

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Thus, at y = 1, the line meets the y-axis.

Hence, the initial value of the line is: y = 1

Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the y-intercept of y = 2x+1 is y = 1.

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Therefore, the y-intercept of y = 2x+1 is y = 1.

Conclusion:

Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

6 0
3 years ago
CALCULUS: Determine which function is a solution to the differential equation y ' − y = 0.
Montano1993 [528]

C: none of these are solutions to the given equation.

• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.

• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.

The actual solution is easy to find, since this equation is separable.

<em>y'</em> - <em>y</em> = 0

d<em>y</em>/d<em>x</em> = <em>y</em>

d<em>y</em>/<em>y</em> = d<em>x</em>

∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>

ln|<em>y</em>| = <em>x</em> + <em>C</em>

<em>y</em> = exp(<em>x</em> + <em>C </em>)

<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>

8 0
3 years ago
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