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2 years ago

PLEASE , ANYONE SAY THE ANSWER!! PLEASE I NEED IT , PLS ANYONE HELP

Mathematics

1 answer:

klasskru [66]2 years ago

7 0

Answer:

<h2>B. 20% because he only got 400 back </h2>

Pls Mark me as brainliest 

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[Quick Answer Needed] Which of the following shows the extraneous solution to the logarithmic equation?

Nitella [24]

Answer:

Step-by-step explanation:

Given the logarithmic equation

$\log_4x+\log_4(x-3)=\log_4(-7x+21)$

First, notice that

$x>0\\ \\x-3>0\Rightarrow x>3\\ \\-7x+21>0\Rightarrow 7x$

So, there is no possible solutions, all possible solutions will be extraneous.

Solve the equation:

$\log_4x+\log_4(x-3)=\log_4x(x-3),$

then

$\log_4x(x-3)=\log_4(-7x+21)\\ \\x(x-3)=-7x+21\\ \\x^2-3x+7x-21=0\\ \\x^2+4x-21=0\\ \\D=4^2-4\cdot 1\cdot (-21)=16+84=100\\ \\x_{1,2}=\dfrac{-4\pm 10}{2}=-7,\ 3$

Hence, $x=3$ and $x=-7$ are extraneous solutions

6 0

2 years ago

Factorise completely 12 x³y - 18 xy²

jarptica [38.1K]

Answer:

6xy(2x²-3y)

Step-by-step explanation:

factorized completely

6 0

2 years ago

In ΔJKL, k = 9.6 cm, l = 2.7 cm and ∠J=43°. Find ∠L, to the nearest 10th of a degree.

Bogdan [553]

Answer:

L = 10.64°

Step-by-step explanation:

From the given information:

In triangle JKL;

line k = 9.6 cm

line l = 2.7 cm; &

angle J = 43°

we are to find angle L = ???

We can use the sine rule to determine angle L:

i.e

$\dfrac{j}{SIn \ J} = \dfrac{l}{ SIn \ L}$

Using Pythagoras rule to find j

i,e

j² = k² + l²

j² = 9.6²+ 2.7²

j² = 92.16 + 7.29

j² = 99.45

$j = \sqrt{99.45}$

j = 9.97

∴

$\dfrac{9.97}{Sin \ 43} = \dfrac{2.7}{ Sin \ L}$

${9.97 \times Sin (L ) = (2.7 \times Sin \ 43)$

$= Sin \ L = \dfrac{ (2.7 \times Sin \ 43)}{9.97 } \\ \\ = Sin \ L = \dfrac{ (2.7 \times 0.6819)}{9.97 } \\ \\ = Sin \ L = 0.18466 \\ \\ L = Sin^{-1} (0.18466) \\ \\ L = 10.64 ^0$