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2 years ago

Can someone plz help me with this one problem

Mathematics

2 answers:

bulgar [2K]2 years ago

8 0

Answer:

Yes

Step-by-step explanation:

I´m pretty sure but not 100% sure

sveticcg [70]2 years ago

4 0

Yes could be wrong sorry if it is

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Please help!! I’m stuck on this

beks73 [17]

Answer:

B. $x = -4\,\lor \,x = -1$

Step-by-step explanation:

Let $\sqrt{-4-5\cdot x} = x$ , we solve for $x$ by algebraic means.

1) $\sqrt{-4-5\cdot x} = x$ Given

2) $-4-5\cdot x = x^{2}$ Definition of power/Symmetry of equality

3) $x^{2}+5\cdot x + 4 = 0$ Compatibility with addition/Commutative and associative properties/Existence of additive inverse/Modulative property

4) $(x+4)\cdot (x+1) = 0$ $x^{2}+(a+b)\cdot x +a\cdot b = (x + a)\cdot (x+b)$

5) $x = -4\,\lor \,x = -1$ Result

Hence, the correct answer is B.

7 0

2 years ago

Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.$

3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$