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Shalnov [3]
3 years ago
12

Write the expression without radicals, using only positive exponents

Mathematics
2 answers:
Mamont248 [21]3 years ago
7 0

Answer:

\sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}

Step-by-step explanation:

∵∛x = (x)^1/3

∴ \sqrt[3]{a^{2}+b^{2}}=(a^{2}+b^{2})^{\frac{1}{3}}

So you can replace the radicals by fractional exponents

marusya05 [52]3 years ago
4 0

Answer:

The required expression is: \sqrt[3]{a^2+b^2}=(a^2+b^2)^{\frac{1}{3}.

Step-by-step explanation:

Consider the provided expression.

\sqrt[3]{a^2+b^2}

Now use the property: \sqrt[n]{x+y}=(x+y)^{\frac{1}{n}}

By using the above property the provided expression can be written as:

\sqrt[3]{a^2+b^2}=(a^2+b^2)^{\frac{1}{3}

Therefore, the required expression is: \sqrt[3]{a^2+b^2}=(a^2+b^2)^{\frac{1}{3}.

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Two collinear points on a line are given in the table below:
katrin [286]

Answer:

(4,3) and (7,2) do not lie on the line

Step-by-step explanation:

Given

(0,0)\ and\ (2,1)

Required

Determine which points that are not on the line

First, we need to determine the slope (m) of the line:

m = \frac{y_2 - y_1}{x_2- x_1}

Where

(x_1,y_1) = (0,0)

(x_2,y_2) = (2,1)

So;

m = \frac{y_2 - y_1}{x_2- x_1}

m = \frac{1 - 0}{2-0}

m = \frac{1}{2}

Next, we determine the line equation using:

y - y_1 = m(x -x_1)

Where

m = \frac{1}{2}

(x_1,y_1) = (0,0)

y - y_1 = m(x -x_1) becomes

y - 0 = \frac{1}{2}(x - 0)

y = \frac{1}{2}x

To determine which point is on the line, we simply plug in the  values of x to in the equation check.

For (4,2)

x = 4 and y =2

Substitute 4 for x and 2 for y in y = \frac{1}{2}x

2 = \frac{1}{2} * 4

2 = \frac{4}{2}

2=2

<em>This point is on the graph</em>

<em></em>

For (4,3)

x = 4 and y = 3

Substitute 4 for x and 3 for y in y = \frac{1}{2}x

3 = \frac{1}{2} * 4

3 = \frac{4}{2}

3 \neq 2

<em>This point is not on the graph</em>

<em></em>

For (7,2)

x = 7 and y = 2

Substitute 7 for x and 2 for y in y = \frac{1}{2}x

2 = \frac{1}{2} * 7

2 = \frac{7}{2}

2 \neq 3.5

<em></em>

<em>This point is not on the graph</em>

<em></em>

<em></em>(\frac{4}{8},\frac{2}{8})<em></em>

<em></em>x = \frac{4}{8} and<em> </em>y = \frac{2}{8}<em></em>

<em>Substitute </em>\frac{4}{8}<em> for x and </em>\frac{2}{8}<em> for y in </em>y = \frac{1}{2}x<em></em>

<em></em>\frac{2}{8} = \frac{1}{2} * \frac{4}{8}<em></em>

<em></em>\frac{2}{8} = \frac{1 * 4}{8 * 2}<em></em>

<em></em>\frac{2}{8} = \frac{4}{16}<em></em>

<em></em>\frac{1}{4} = \frac{1}{4}<em></em>

<em></em>

<em>This point is on the graph</em>

3 0
3 years ago
Using the Triangle Sum Theorem, find x and the missing angles (2x +1) (5x +5) x= Angle (2x+1) = Angle (5x+5) =​
Arte-miy333 [17]

Answer:

x = 12

(2x + 1)° = 25°

(5x + 5)° = 65°

Step-by-step explanation:

By the property of a triangle,

Sum of all angles of a triangle is 180°

In the given right triangle,

(2x + 1)° + 90° + (5x + 5)°= 180°

7x + 96 = 180

7x = 180 - 96

x = \frac{84}{7}

x = 12

Angle (2x + 1)° = 2(12) + 1 = 25°

Angle (5x + 5)° = 5(12) + 5 = 65°

Therefore, all three angles of the right triangle are 25°, 90° and 65°.

4 0
3 years ago
What is x<br>-5(x-2)-(x-2)=50
stich3 [128]
X= -14/3 that's the answer there bud i need to make this 20 caracters longs so yeah x=-14/3
4 0
3 years ago
Hdndnndndndndnndndndndnbd
ipn [44]

Answer:

Hshushahhahaa

Hdhddjkddjdkjd

8 0
3 years ago
What is the solution of log4(2x-6)=2
Flauer [41]

Answer: x=11

Step-by-step explanation:

Remembert that, by definition:

log_b(x)=y → b^y=x

Then, you can rewrite log_4(2x-6)=2 in exponential form:

4^2=2x-6

Now you can solve for the variable "x":

Add 6 to both sides of the equation:

4^2+6=2x-6+6

22=2x

And finally you must divide both sides of the equation by 2, then:

\frac{22}{2}=\frac{2x}{2}\\\\x=11

3 0
3 years ago
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