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4 years ago

BRAINLIEST WILL BE MARKED

Mathematics

1 answer:

zubka84 [21]4 years ago

8 0

Answer:

B. input

Step-by-step explanation:

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The endpoints of a side of rectangle ABCD in the coordinate plane are at A(1,4) and

andrew-mc [135]

Answer:

y = - x + 5

Step-by-step explanation:

A(1, 4)

B(4, 1)

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3 years ago

Factor:

nekit [7.7K]

Answer:

a) $(7\cdot m^{6}+4)\cdot (7\cdot m^{6}-4)$

b) $(7\cdot a\cdot b^{8}+1)\cdot (7\cdot a\cdot b^{8}-1)$

c) $a = 0.7\cdot t^{9}$

d) There are two possible answers only considering real coefficients:

(i) $16\cdot x^{2}-81 = (4\cdot x +9)\cdot (4\cdot x - 9)$

(ii) $14\cdot x^{2}-81 =(\sqrt{14}\cdot x+9)\cdot (\sqrt{14}\cdot x - 9)$

e) $(a^{3}+4)\cdot (a^{3}-4)$

Step-by-step explanation:

Now we proceed to solve each algebraic equation:

a) Factor $49\cdot m^{12}-16$

This binomial is of the form $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . In this case, we can rewrite and factor the equation below:

$49\cdot m ^{12}-16$

$(7\cdot m^{6})^ 2-4^{2}$

$(7\cdot m^{6}+4)\cdot (7\cdot m^{6}-4)$

b) Factor $49\cdot a^{2}\cdot b^{16}-1$

This binomial is of the form $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . In this case, we can rewrite and factor the equation below:

$49\cdot a^{2}\cdot b^{16}-1$

$(7\cdot a\cdot b^{8})^{2}-1^{2}$

$(7\cdot a\cdot b^{8}+1)\cdot (7\cdot a\cdot b^{8}-1)$

c) Nicole is factoring $0.49\cdot t^{18}-25$ by using the rule $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . What will she use for the value of $a$ ?

From this rule we find that $a^{2} = 0.49\cdot t^{18}$ , that is:

$a^{2} = 0.49\cdot t^{18}$

$a^{2} = \frac{49}{100}\cdot t^{18}$

$a = \frac{7}{10}\cdot t^{9}$

$a = 0.7\cdot t^{9}$

d) Which binomial is a difference of squares?

A binomial is a difference of square if and only if $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . There are two possible answers only considering real coefficients:

(i) $16\cdot x^{2}-81 = (4\cdot x +9)\cdot (4\cdot x - 9)$

(ii) $14\cdot x^{2}-81 =(\sqrt{14}\cdot x+9)\cdot (\sqrt{14}\cdot x - 9)$

e) Factor $a^{6}-16$

This binomial is of the form $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . In this case, we can rewrite and factor the equation below:

$a^{6}-16$

$(a^{3})^{2}-4^{2}$

$(a^{3}+4)\cdot (a^{3}-4)$

6 0

3 years ago