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Tanzania [10]
3 years ago
10

What is the approximate length of arc QR

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
4 0
What arc QR?

Is there supposed to be a picture here?
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Evaluate the expression for the given value of the variable. <br><br> 12 – 3b; b = 4.3
vladimir1956 [14]

Answer:

24.9

Step-by-step explanation:

12-3b    b=4.3

12-3(4.3)

12- (-12.9)

12 + 12.9

24.9

3 0
4 years ago
-1/2x + 3 - 4x + 3/2x
SashulF [63]

Answer:

-6 x (X-1)/2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

                                           =x^2h

Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

=20x^2 cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

=4xh\times 1.5 cents

=6xh cents

Total cost = (20x^2+10x^2+6xh)

                =30x^2+6xh

Let

C=30x^2+6xh

Putting the value of h

C=30x^2+6x\times \frac{272}{x^2}

\Rightarrow C=30x^2+\frac{1632}{x}

Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

                                          =30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0
3 years ago
Write the formula for an exponential function with initial value 300 and decay factor 0.73
postnew [5]

Answer: 300*0.73^x

Step-by-step explanation:

300 is the initial value and every time x goes up by 1, the value get multiplied by 0.73

when x=3

then the value would be 300*0.73*0.73*0.73 or 300*0.73^3

then for x amount of times the equation would be 300*(0.73 x times ) or 300*0.73^x

8 0
3 years ago
Simplify the expression.<br><br> 2/3 × [15 × 5 + (-24)] + (-18)
prisoha [69]
<span>2/3 x [ 15 x 5 + ( -24 )] + ( -18 ) =  2/3 * [ 15 * 5 - 24 ] - 18 = 

= 2/3 * [ 45 - 24 ] - 18 = 2/3 * 21 - 18 = 2/3 * 21/1 - 18 = 2/1 * 7/1 - 18 =

= 14 - 18 = -4


</span>
4 0
3 years ago
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