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3 years ago

4.7j-6.8k when j =5 and k =3

Mathematics

1 answer:

Oksanka [162]3 years ago

5 0

Answer:

Your answer is 3.1

Step-by-step explanation:

Wherever you see j, substitute 5. Wherever you see k, substitute 3

1. Set up: $4.7(5)-6.8(3)$

2. Multiply 4.7 * 5, to get 23.5

3. Multiply 6.8 * 3 to get 20.4

4. Subtract 23.5-20.4

5. Final anwer - 3.1

Hope this helps!

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What type of muffin coast the least?

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type C costs the least per muffin

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3 years ago

SinA/sinB=2/√3 and cosA/cosB=1/√2 ,tan^2B=?

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2 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
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Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
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Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
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Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
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Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
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3 years ago

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Step-by-step explanation:

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3 years ago

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A dog is leashed to the corner of a house with a 20 ft long leash. How much running area does the dog have? Round your answer to

11111nata11111 [884]

Answer:

The dog's running area is approximately 942 feet².

Step-by-step explanation:

The dog is leashed to a fixed point, the leash has a length of 20 ft, therefore he can rotate around that point at the maximum distance equal to the length of the leash. This pattern forms a circle, but there is an obstruction, which is the corner of the house. This obstruction takes an arc of the original circle, so the running area of the dog is the area of the whole circle minus the area of the arc formed by the corner of the house.

$\text{dog's area} = \text{circle's area} - \text{arc's area}\\\\\text{dog's area} = pi*(20^2) - \frac{90}{360}*pi*(20^2)\\\\\text{dog's area} = \frac{270}{360}*pi*(400)\\\\\text{dog's area} = 942.477\text{ feet}^2$

The dog's running area is approximately 942 feet².

3 0

3 years ago