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3 years ago

Answer true or false for 1-10 (picture above)

Mathematics

1 answer:

Softa [21]3 years ago

3 0

$\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}\\\\\sqrt{a-b}\geq\sqrt{a}-\sqrt{b}\\----------------------\\1)\ \sqrt{72}=\sqrt{9\cdot8}=\sqrt{9}\cdot\sqrt{8}\qquad TRUE\\\\2)\ \sqrt{13-5}=\sqrt{13}-\sqrt5\qquad FALSE\\\\3)\ \sqrt{9\cdot10}=\sqrt9+\sqrt{10}\qquad FALSE\\\\4)\ \sqrt{18bx}=\sqrt{18}\cdot\sqrt{b}\cdot\sqrt{x}\qquad TRUE\ if\ b\geq0\ and\ x\geq0\\\\5)\ \sqrt{8\cdot12}=\sqrt8\cdot\sqrt{12}\qquad TRUE\\\\6)\ \sqrt{m+x}=\sqrt{m}+\sqrt{x}\qquad FALSE$

$7)\ \sqrt{100+64}=\sqrt{100}+\sqrt{64}\qquad FALSE\\\\8)\ \sqrt{30}=\sqrt{5\cdot6}=\sqrt5\cdot\sqrt6\qquad TRUE\\\\9)\ \sqrt{44}=\sqrt{22+22}=\sqrt{22}+\sqrt{22}\qquad FALSE\\\\10)\ \sqrt{10\cdot10}=\sqrt{10}\cdot\sqrt{10}\qquad TRUE$

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How many solutions does the system have? y = -2x-4 \\\\ y = 3x+3

lakkis [162]

Answer:

The system has one solution.

Step-by-step explanation:

We have two equations:

y = -2x - 4

y = 3x + 3

Equalling them:

y = y

-2x - 4 = 3x + 3

5x = -7

$x = -\frac{7}{5}$

And

$y = 3x + 3 = 3(-\frac{7}{5}) + 3 = \frac{-21}{5} + 3 = \frac{-21}{5} + \frac{15}{5} = -\frac{6}{5}$

Replacing in the other equation we should get the same result.

$y = -2x - 4 = -2(-\frac{7}{5}) - 4 = \frac{14}[5} - 4 = \frac{14}{4} - \frac{20}{5} = -\frac{6}{5}$

So the system has one solution.

3 0

3 years ago

Hhheleppppp urgggeenttt

tekilochka [14]

They intercept at (-1,5)

6 0

3 years ago

-cep Leu<br>2. 3x -1 = -10

nekit [7.7K]

Answer:

-3

Step-by-step explanation:

3x-1=-10

3x=-9

x=-3

7 0

3 years ago

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

6 0

3 years ago

1.9.2) If the programme started at 12:30 pm, at what time will it end?

DerKrebs [107]

Answer:

We need to know what the program is...

Step-by-step explanation:

This question is literally unanswerable as some programs last 2 seconds and some would last thousands of years. More context as to what the program is please.

3 0

3 years ago