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2 years ago

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

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2 years ago

In relation defined by the equation y=2x+1 for all x>0, y is a function because of x because

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that would be the letter d :)

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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

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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