All categories

3 years ago

Please help. will mark brainliest

Mathematics

1 answer:

jeka943 years ago

6 0

2x/2 = 0/2

x = 0

2x/2 = 54/2

x = 27

0 < x < 27

C. 0 < x < 27

You might be interested in

Which equation below gives an incorrect value for the function k(x) = 512x?

lina2011 [118]

Answer:

D. gives an incorrect value for the function k(x) = $512^{x}$

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

Which equation below gives an incorrect value for the function k(x) = $512^{x}$ ?

A.) k(2/3)=64

B.) k(4/9)=16

C.) k(1/3)=8

D.) k(2/9)=8

My answer:

Let analyse all 4 possible answers:

A.) k(2/3)=64

<=> k(2/3) = $512^{2/3}$ = 64

True

B.) k(4/9)=16

<=> k(4/9) = $512^{4/9}$ = 16

True

C. k(1/3)=8

<=> k(1/3)= $512^{1/3}$ = 8

True

D. k(2/9)=8

<=> k(2/9) = $512^{2/9}$ = 4 not 8

Wrong

Hope it will find you well

3 0

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

2 years ago

Ax+by=(a-b) , bx - ay =(a+b) simultaneous linear equations using cross multiplication

Molodets [167]

Answer:

$x=1\,,\,y=-1$

Step-by-step explanation:

Given: $ax+by=a-b\,,\,bx-ay=a+b$

To solve: the given linear equations

Solution:

Consider the equations:

$A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0$

By method of cross multiplication:

$\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-C_2A_1}=\frac{1}{A_1B_2-A_2B_1}$

For equations: $ax+by=a-b\,,\,bx-ay=a+b$

$ax+by-(a-b)=0\\bx-ay-(a+b)=0$

Take $A_1=a\,,\,B_1=b\,,\,C_1=-(a-b)\,,\,A_2=b\,,\,B_2=-a\,,\,C_2=-(a+b)$

So,

$\frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^2-b^2}\\\frac{x}{-ab-b^2-a^2+ab}=\frac{y}{-ab+b^2+a^2+ab}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}\,,\,\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\x=1\,,\,y=-1$