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Cloud [144]
3 years ago
5

What is 528−−√+63−−√528+63 in simplest radical form?

Mathematics
2 answers:
Montano1993 [528]3 years ago
6 0

Answer: 13\sqrt{7}

Step-by-step explanation:

In the simple radical form there are no square root is remain to find.

Since, the given expression,

5\sqrt{28} + \sqrt{63}

5\sqrt{4\times 7} + \sqrt{9\times 7}

5\sqrt{4} \times \sqrt{7} + \sqrt{9}\times \sqrt{7}

5\times 2 \times \sqrt{7} + 3\times \sqrt{7}

10 \sqrt{7} + 3\sqrt{7}

(10 + 3)\sqrt{7}

13\sqrt{7}

Since, we do not need to find further square root of 7.

Thus, the required radical form of 5\sqrt{28} + \sqrt{63} is 13\sqrt{7}.

Rufina [12.5K]3 years ago
3 0

Answer:

13\sqrt{7}

Step-by-step explanation:

5\sqrt{28} +\sqrt{63}

To simplify the given expression we simplify each radical

5\sqrt{28} = 5\sqrt{4*7} = 5*2\sqrt{7} =10\sqrt{7}

\sqrt{63} = \sqrt{9*7} =3\sqrt{7}

5\sqrt{28} +\sqrt{63}

10\sqrt{7}+3\sqrt{7}

13\sqrt{7}

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What is the equation of the perpendicular line 2x-4y=7 passing through the points (2,-5/4)?
RUDIKE [14]

Answer:

y= 2x-5\frac{1}{4}

Step-by-step explanation:

We are given;

The equation 2x + 4y = 7

A point (2, -5/4)

We are supposed to determine the equation of a line perpendicular to the given line;

First, we determine the slope of the given line by writing the equation in the form of y= mx + c

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m₁ = -1/2

But, for perpendicular; m₁×m₂= -1

Therefore;

-1/2 × m₂ = -1

m₂ = 2

Thus, we can get the equation of the line;

Taking another point (x, y)

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y+\frac{5}{4}= 2(x-2)

y+\frac{5}{4}= 2x-4

y= 2x-4-\frac{5}{4}

y= 2x-5\frac{1}{4}

Thus, the equation of the line in question is y= 2x-5\frac{1}{4}

5 0
4 years ago
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k0ka [10]

Answer:

m = 2.25

b = 4.15

Step-by-step explanation:

Let us consider the points (1, 6.4) & (3, 10.9)

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Now, by point slope formula of a line, we have:

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\therefore y - 6.4 = 2.25x - 2.25

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\therefore y = 2.25x + 4.15

Equating it with y = mx + b, we find:

m = 2.25

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azamat

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Step-by-step explanation:

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