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3 years ago

Which expression is equivalent to 4 square root Of x^10

Mathematics

1 answer:

Airida [17]3 years ago

4 0

A is the answer to this question

You might be interested in

1st question: is 3×2/3 greater or less than 3 ........ 2nd question: is 5/7×3 greater or less than 5/7?

Dominik [7]

First one:
3 * 2/3 is equal to 6/3 which is equal to 2 and 2 is less than 3
second one:
5/7 * 3 is equal to 15/7 and 15/7 is greater than 5/7 :)))
i hope this is helpful
have a nice day

4 0

3 years ago

What is the length of segment DB?

tatyana61 [14]

I believe it’s 82

8x=3x+8+4x+10
x=18

4•18+10=72+10=82

3 0

3 years ago

Find 2 consecutive positive even integers such that the product of the 1st and 2nd is 2 less than 5 times the 3rd. Use an algebr

Aloiza [94]

Assume the smaller one is x, then the other ones will be x+2 and x+4

x(x+2) + 2 = 5(x+4)

x^2 + 2x - 5x - 18 = 0

-> x^2 - 3x - 18 = 0

-> (x-6)(x+3) = 0

-> x = 6 or -3, since x must be positive, then x = 6

So the numbers are: 6, 8, 10

6 0

4 years ago

To find the area of a triangle, you can use the expression b × h ÷ 2, where b is the base of the triangle and h is its height. W

Zolol [24]

Answer:

Step-by-step explanation:

b x h is finding a square or rectangle, but since a triangle is half of a rectangle or square you can divide that by 2. So, 8 (b) x 4 (h) = 32, 32 / 2 = 16. Hope this helps! :D

6 0

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

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

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

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

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

3 0

3 years ago