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2 years ago

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

Mathematics

2 answers:

Montano1993 [528]2 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]2 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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Answer:

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Step-by-step explanation:

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2 years ago

Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter of the so

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In order to construct this equation, we will use the variables:
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We know that the total amount of substance, product of the volume and concentration, in the final solution is equal to the individual amounts in the two given solutions. Thus:
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2 years ago

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Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. Be sure to create and name

sattari [20]

Answer:

Given: A triangle ABC and a line DE parallel to BC.

To prove: A line parallel to one side of a triangle divides the other two sides proportionally.

Proof: Consider ΔABC and DE be the line parallel to Bc, then from ΔABC and ΔADE, we have

∠A=∠A (Common)

∠ADE=∠ABC (Corresponding angles)

Thus, by AA similarity, ΔABC is similar to ΔADE, therefore

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2 years ago

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2 years ago

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

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