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Mamont248 [21]
3 years ago
10

Square root of 10x^2 times square root of 5x

Mathematics
1 answer:
Deffense [45]3 years ago
6 0

Answer:

5√2x2

Step-by-step explanation:

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Match each simplified form to it's equivalent radical expression. Tiles: 4√3, 3 3√2, 2 3√3, 3√5
lbvjy [14]

Answer:

Step-by-step explanation:

Given the following simplified expressions:

4√3, 3 3√2, 2 3√3, 3√5

It's radical equivalent is :

4√3 = √4² * √3 = √16 * √3 = √16*3 = √48

3 3√2 = 3 * √3² * √2 = √9 * √2 = √9*2 = 3√18

2 3√3 = 2 * √3² * √3 = √9 * √3 = √9*3 = 2√27

3√5 = √3² * √5 = √9 * √5 = √9*5 = √45

4 0
3 years ago
Jim’s allowance is $1.20 per week. Stan’s is 25¢ per day. How long will they have to
algol [13]
First, find out how much stan gets per week. 

.25 x 7 =1.75

now how much they get together

1.75 + 1.20 = 2.95

and then the final step

23.60 / 2.95 = 8

it'll take 8 weeks for them to buy the model car set
7 0
2 years ago
Can someone please help me
Ainat [17]

Answer:

20

Step-by-step explanation:

If the two triangles are similar, then corresponding sides must share a constant ratio. This means that:

\dfrac{10}{6}=\dfrac{25}{15}=\dfrac{x}{12}

Let's use the second ratio:

\dfrac{25}{15}=\dfrac{x}{12}

Multiply both sides by 12:

\dfrac{25\cdot 12}{15}=x \\\\x=20

Hope this helps!

5 0
2 years ago
If 70 was removed from the set of data, which values would change?
AURORKA [14]
Its mean and range . Even if we remove 70 the median is the same. The mean changes also as 580/7 isnt the same as 510/6 and also the range changes as it decreases by 5. Hope it helped :)
6 0
2 years ago
How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

8 0
3 years ago
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