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DiKsa [7]
3 years ago
14

Ribbon ones are made from strips of ribbon tied to sticks Connie has 84 feet of red ribbon 48 feet of blue ribbon and 72 feet of

white ribbon she wants to cut the ribbon into equal links that are as long as possible so that no ribbon is wasted
Mathematics
1 answer:
WARRIOR [948]3 years ago
4 0
Red: 7 12-foot pieces
blue: 4 12-foot pieces
white: 6 12-foot pieces
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If the area of the rectangle is 42 square units, what is the area of the shaded part?
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In a group of 90 seventh
Firdavs [7]
90 x .80= 72. 72 Students like hot choco, therefore 8 do no like hot choco
5 0
2 years ago
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A triangle is shown.
aleksklad [387]

Using the Pythagorean Theorem, it is found that the unknown side length of the triangle is of 7.9 cm.

<h3>What is the Pythagorean Theorem?</h3>

The Pythagorean Theorem relates the length of the legs l_1 and l_2 of a right triangle with the length of the hypotenuse h, according to the following equation:

h^2 = l_1^2 + l_2^2

Researching this problem on the internet, we have that:

  • The unknown side length is of a leg of x.
  • The other leg is of 9 cm.
  • The hypotenuse is of 12 cm.

Hence:

x² + 9² = 12²

x = sqrt(12² - 9²)

x = 7.9 cm.

The unknown side length of the triangle is of 7.9 cm.

More can be learned about the Pythagorean Theorem at brainly.com/question/654982

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4 0
2 years ago
NEED QUICKLY <br> Write the equation of the line shown below?
grandymaker [24]

y = x + 2

Step-by-step explanation:

General formula for any straight line:

y = mx + c

Where

m = gradient

c = constant

m = y2 - y1/x2 - x1

m = (4 - ( -2))/(2 - ( -4))

m = 6/6

m = 1

y = (1)x + c

y = x + c

Substitute any coordinate from the line of the equation.

4 = 2 + c

c = 2

substitute m and c into general formula

y = mx + c

y = x + 2

8 0
2 years ago
Read 2 more answers
Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
Tema [17]

Answer:

The area can be written as

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

  • x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
  • x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

x_u = 1/v

x_v = u*ln(v)

y_u = v

y_v = u

Therefore, the Jacobian matrix is

\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}

Therefore,

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274

4 0
3 years ago
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