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creativ13 [48]
3 years ago
15

Which equation can be used to solve for x in the following diagram?

Mathematics
2 answers:
mestny [16]3 years ago
7 0
C
hope this helps you
Butoxors [25]3 years ago
4 0

Answer:

The answer would be C.

Step-by-step explanation:

You can tell that the triangle is a right angle by the box in the corner which makes it 90 degrees so 7x+11x would equal 90

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The area of a rectangle is 30 square feet. The base of the rectangle is 7.5
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4 ft

Step-by-step explanation:

<h3>Area of rectangle:</h3>

Area = 30 square ft

base = 7.5 ft

          \sf \boxed{height \ of \ rectangle = \dfrac{Area}{base}}

                                             \sf =\dfrac{30}{7.5}\\\\=\dfrac{30*10}{7.5*10}\\\\=\dfrac{300}{75}\\\\= 4 ft

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Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 &lt; = x &lt; = 8. The length of the curve is . (Type an exact an
Mashutka [201]

Answer:

\sqrt\frac{387}{20}

Step-by-step explanation:

Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx

y=\dfrac{3}{5}x^{\frac{5}{3}}-  \dfrac{3}{4}x^{\frac{1}{3}}+6

\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}

1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})

=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}

For the Interval 1\leq x\leq 8

Length of the Curve =\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\

Using T1-Calculator

=\sqrt\frac{387}{20}

3 0
4 years ago
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