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2 years ago

A sandbox has a area of 26 square feet and the length is 5 1/2 feet what is the width of the sandbox?

Mathematics

2 answers:

nexus9112 [7]2 years ago

7 0

We can write an equation

26=5.5*x

26 is the area and 5.5 is the length

To get x alone divide26 by 5.5

The answer is 52/11

Ulleksa [173]2 years ago

3 0

Answer:

4.5 ft

Step-by-step explanation:

Area = length times width. Let A = 26 ft^2 and divide it by L = 5.5 ft, to obtain the width W. W = 26 ft^2 / 5.5 ft = 4.54 ft, or 4.5 ft to the nearest tenth foot.

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M(x) = -2x. Find m(5).<br>m(5) =

Natasha_Volkova [10]

Answer:

this is the correct answer look this img

6 0

2 years ago

-5x-3.5 < 6.5<br> -5x <10<br> x >2

Studentka2010 [4]

Answer:

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Step-by-step explanation:

cuz it makes sense like that

6 0

2 years ago

You are attempting to estimate the number of protester marching through the city

Alenkinab [10]

9514 1404 393

Answer:

1000

Step-by-step explanation:

If the number of protesters per minute remains a constant, then you could write the proportion ...

p/12 = 177/2.1

Multiplying by 12 gives ...

p = 12(177/2.1) ≈ 1011.4

Here, minutes are given to 2 significant figures, and the initial count is given to 3 significant figures. The best you can hope for is that your estimate is good to 3 significant figures:

1010 protesters

It is probably sufficient to report the number to 2 significant figures*:

1000 protesters

_____

* Unfortunately, with a number like 1000, the only way you can tell it has 2 significant figures is to report it as 1.0×10³ or 10. hundreds. The trailing zeros are usually not considered significant.

3 0

2 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$