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2 years ago

What is the following product? (2V7 +3V6)(5V2+4V3)

Mathematics

1 answer:

raketka [301]2 years ago

5 0

$(2\sqrt{7}+3\sqrt{6})(5\sqrt{2}+4\sqrt{3}) \\\\\\ \begin{array}{lllllll} 10\sqrt{14}&+&15\sqrt{12}&+&8\sqrt{21}&+&12\sqrt{18}\\\\ &&15\sqrt{2^2\cdot 3}&&&&12\sqrt{3^2\cdot 2}\\\\ &&15\cdot 2\sqrt{3}&&&&12\cdot 3\sqrt{2}\\\\ &&30\sqrt{3}&&&&36\sqrt{2}\\\\ 10\sqrt{14}&+&30\sqrt{3}&+&8\sqrt{21}&+&36\sqrt{2}\\\\ 10\sqrt{14}&+&8\sqrt{21}&+&30\sqrt{3}&+&36\sqrt{2} \end{array}$

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Expression.<br><br> 3(4x + 3)

SVETLANKA909090 [29]

3(4x+3) simplified as an expression is: 12x+9

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3 years ago

Pl help it’s for a grade

allsm [11]

Answer:

(^6√x^5) (√y)

(The 6 belongs in the left on top of the square root btw)

For x the 6 moves to the front of the square root while the 5 becomes the power of x.

For y, the power to the 1/2 is the same as a square root.

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3 years ago

Answer the questions below about the quadratic function.

Korolek [52]

Given the function $f(x)=3x^2-18x+23$ . The above function can be written as

$f(x)=3x^2-18x+23\\ f(x)=3(x^2-6x)+23\\ f(x)=3(x^2-6x+9-9)+23\\ f(x)=3(x-3)^2-27+23\\ f(x)=3(x-3)^2-4$

a)Now, the function $f(x)=3(x-3)^2-4$ has minimum value since the coefficient of $(x-3)^2$ is $3>0$ .

b) The minimum value of the function occurs at $x=3$ and its value is

$f(3)=3(3-3)^2-4 =-4$

c)The minimum value of the function occurs at $x=3$ .

6 0

3 years ago

PLASE ANSWER THIS QUESTION I GIVE YOU 18 POINTS

sukhopar [10]

Answer:

(a) 144

(b) 117

(d) 588

(e) 5472

Step-by-step explanation:

To find the LCM of two numbers, first find the prime factorization of each number. Then the LCM is the product of common and not common factors with the larger exponent.

(a)

$72 = 2^3 \times 3^2$

$144 = 2^4 \times 3^2$

$LCM = 2^4 \times 3^2 = 8 \times 9 = 144$

(b)

$39 = 3 \times 13$

$117 = 3^2 \times 13$

$LCM = 3^2 \times 13 = 117$

(c)

$72 = 2^3 \times 3^2$

$90 = 2 \times 3^2 \times 5$

$LCM = 2^3 \times 3^2 \times 5 = 360$

(d)

$84 = 2^2 \times 3 \times 7$

$147 = 3 \times 7^2$

$LCM = 2^2 \times 3 \times 7^2 = 588$

(e)

$152 = 2^3 \times 19$

$288 = 2^5 \times 3^2$

$LCM = 2^5 \times 3^2 \times 19 = 5472$

6 0

3 years ago

In january of one winter, 3/10 of a wood pile was used in a wood stove. In February, 2/5 of the wood pile was used. What part of

arsen [322]

2/5 is equivalent to 4/10 so at the end of February 4/10 was used up
And at the end of January, 3/10 was used
so 4/10 + 3/10 = 7/10

At the end of February, 7/10 of the wood was used

6 0

3 years ago