What is the solution to log Subscript 3 Baseline (x + 12) = log Subscript 3 Baseline (5 x)

GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation −2y=3x+7 in the form $y=-\dfrac{3}{2}x-\dfrac{7}{2}.$ Here the slope of the given line is $m_1=-\dfrac{3}{2}.$ If $m_2$ is the slope of perpendicular line, then

$m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.$

Answer 1: $\dfrac{2}{3}$

Part B. The slope of the line y=−2x+3 is -2. Since $-\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2,$ then lines from part A are not parallel to line a.

Since $-2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1,$ both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line $y=\dfrac{x}{4}+5$ is $\dfrac{1}{4}.$ Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then

7=-4·2+b,

b=7+8,

b=15.

Answer 4: y=-4x+15.

Part E. Consider vectors $\vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a).$ These vectors are collinear, then

$\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.$

Answer 5: $-\dfrac{a}{b}=-\dfrac{d}{c}.$

5 0

3 years ago

Read 2 more answers

Please solve the following sum or difference identity.

xxTIMURxx [149]

Answer:

$sin(A - B) = \frac{4}{5}$

Step-by-step explanation:

Given:

$sin(A) = \frac{24}{25}$

$sin(B) = -\frac{4}{5}$

Need:

$sin(A - B)$

First, let's look at the identities:

sum: $sin(A + B) = sinAcosB + cosAsinB$

difference: $sin(A - B) = sinAcosB - cosAsinB$

The question asks to find sin(A - B); therefore, we need to use the difference identity.

Based on the given information (value and quadrant), we can draw reference triangles to find the simplified values of A and B.

sin(A) = $\frac{24}{25}$

cos(A) = $\frac{7}{25}$

sin(B) = $-\frac{4}{5}$

cos(B) = $\frac{3}{5}$

Plug these values into the difference identity formula.

$sin(A - B) = sinAcosB - cosAsinB$

$sin(A - B) = (\frac{24}{25})(\frac{3}{5}) - (-\frac{4}{5})(\frac{7}{25})$

Multiply.

$sin(A - B) = (\frac{72}{125}) + (\frac{28}{125})$

Add.

$sin(A - B) = \frac{4}{5}$

This is your answer.

Hope this helps!

6 0

3 years ago

Read 2 more answers

How do I do it? I’m get confused what do I type in the box?

erastovalidia [21]

I think that’s called (translations)if i’m not wrong

7 0

3 years ago

There are 10 rolls of film in a box and 3 are defective. two rolls are selected without replacement. what is the probability of

german

P(picking one defective) = 3/10
P(picking a 2nd defective) = 2/9

P(1 and 2 defective) = 3/10 x 2/9 = 6/90 = 0.066

Second method using combination:
³C₂ / ¹⁰C₂ = 1/15 = 0.066

4 0

3 years ago

16x+9=9y-2x Solve for y

Novay_Z [31]

Y=2x+1... I hope this helps love! :)

8 0

3 years ago