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3 years ago

8

Which unit of measure would you use to weigh your bike?

2 answers:

Pavel [41]3 years ago

6 0

Pounds is the correct unit of measure

Y_Kistochka [10]3 years ago

4 0

You would use pounds to weigh your bike

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B. If instead you divide first to solve the equation, what<br> operation would you need to use last?

Sedbober [7]

Answer:

use PEDMAS

P: PARENTHESIS

E: EXPONENTS

D: DIVISON

M: MULTIPLICATION

A: ADDITION

S: SUBTRACTION

Step-by-step explanation:

CAN YOU PLS MARK ME BRAINLIEST THANK YOU !

8 0

3 years ago

Approximate the value of square root of 44 to the nearest hundred

ryzh [129]

The answer is 6.63 Hope this helps!

7 0

3 years ago

Read 2 more answers

Clearly mark on the coordinate plane the location of the ordered pairs and

ladessa [460]

Answer:

Point $J(4,0)$ lies on $x-$ axis and point $M(-2,-3)$ lies quadrant III.

Step-by-step explanation:

A coordinate plane is a plane that is divided into four quadrants by the coordinate axes.

Here, coordinate axes are $x-$ axis and $y-$ axis.

Let's first mark the points in the coordinate plane.

See the attached figure.

From the figure, it can be observed that point $J(4,0)$ lies on $x-$ axis and point $M(-2,-3)$ lies quadrant III.

3 0

3 years ago

81 x 11<br> 73 x82<br> 64 x 55<br> 27 x 63<br> 24 x 25<br> 47 x 63<br> 24 x 44

zubka84 [21]

81 times 11 is 891
hope this helps u figure out your problem.

4 0

3 years ago

Read 2 more answers

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 0

3 years ago