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Virty [35]
3 years ago
14

Claire and 4 friends each bought the same pair of shoes and a purse. The sample shoes did not have a price tag, but the purse wa

s marked $12.95. Their total purchase can be represented by the expression: 5(x + 12.95) where x represents the cost of the shoes. Determine the
simplified expression.

ONE ANSWER FROM BELOW

1. 5x + 12.95
2. -5x - 12.95
3. 5x - 64.75
4. 5x + 64.95
Mathematics
1 answer:
RoseWind [281]3 years ago
3 0

Answer:

Step-by-step explanation:

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Any of the individual surfaces of a solid object pattern of three dimensional shapes.

Step-by-step explanation:

Surface area of a solid object is the measure of the total area of the surface of three dimensional objects. Example a sphere and a cube is a three dimensional objects.

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3 years ago
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Find the measure of the missing angel
labwork [276]

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58 degrees

Step-by-step explanation:

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There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con
grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

\square ABCD

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD

\angle BAC = \angle  DCA

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC

and

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED

So, the congruent sides and angles of △ABE and △CDE are:

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD ---- S

\angle BAC = \angle  DCA ---- A

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED or \[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC  --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0
3 years ago
Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7
rosijanka [135]
\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx

Notice that x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1). Decompose the integrand into partial fractions:

\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}
=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}

Integrating term-by-term, you get

\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx
=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C
5 0
4 years ago
Can someone help me answer this? urgent‼️
Vsevolod [243]
It is proportional because the number are even and can be divided by the same number
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