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JulsSmile [24]
2 years ago
12

Solve for x. Solve for x.

Mathematics
2 answers:
Nitella [24]2 years ago
4 0

Answer:

4√3

Step-by-step explanation:

This is a right triangle. When two sides are given in a right triangle, we need to use the Pythagorean Theorem to solve for the third side.

Pythagorean Theorem: a² + b² = c²

The a here is 11, while the b is x. The c is always the hypotenuse (the longest side), which is 13.

Now we substitute:

11² + x² = 13²

121 + x² = 169

-121           -121

x² = 48

√x² = √48

x = √48 <-- We can simplify that.

√48 = 12 * 4 = 3 * 4 * 4 = 4√3

Svetradugi [14.3K]2 years ago
4 0

Answer:

4*√3

Step-by-step explanation:

By Pythagoras Theorem,

11²+x²=13²

x²=169-121=48

x=4√3

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-x^4y^2+7x^3y^3-3xy^5+2x^2y^4 in descending powers of x
muminat

Answer:

  -x^4y^2 +7x^3y^3 +2x^2y^4 -3xy^5

Step-by-step explanation:

The powers of x in the terms of the given expression are ...

  4, 3, 1, 2

so, we want to swap the last two terms to put them in the desired order:

  -x^4y^2 +7x^3y^3 +2x^2y^4 -3xy^5

6 0
2 years ago
In relation defined by the equation y=2x+1 for all x&gt;0, y is a function because of x because
nekit [7.7K]

that would be the letter d :)

each value of x has a unique value of y

4 0
3 years ago
Which number represents a rational repeating decimal? A. 0.77678 B. 2 5/8 <br> C. 7/9 D. 6.0
hjlf
The answer to your question is A.
3 0
3 years ago
A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs
LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

h=\frac{400}{\pi r^2}

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: h=\frac{400}{\pi r^2}

Therefore:

C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})

C(r)=0.12\pi r^2+\frac{24}{r}

C(r)=\frac{0.12\pi r^3+24}{r}

The minimum cost occurs when the derivative of the Cost =0.

C^{'}(r)=\frac{6\pi r^3-600}{25r^2}

6\pi r^3-600=0

6\pi r^3=600

\pi r^3=100

r^3=\frac{100}{\pi}

r^3=31.83

r=3.17 cm

Recall that:

h=\frac{400}{\pi r^2}

h=\frac{400}{\pi *3.17^2}

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0
3 years ago
3) g(x)= x3 + x<br> h(x) = x + 4<br> Find g(0)h(0)
Arte-miy333 [17]

Answer:

goh(x)=x^3+12x^2+49x+68

Step-by-step explanation:

g(x) = x^3+x

h(x)=x+4

We have to find

goh(x)

goh(x) = g(h(x))=g(x+4)

If g(x) = x^3+x

g(x+4) = (x+4)^3+(x+4)

           =x^3+12x^2+49x+68

Hence our answer is

goh(x)=x^3+12x^2+49x+68

4 0
3 years ago
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