3 years ago

What is the equation of the line that is perpendicular to y = one-fifth x + 4 and that passes through (5,–4)?

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

8 0

Answer:

y = -5x + 21

Step-by-step explanation:

perpendicular lines have slopes that are negative reciprocals.

The given line has a slope of 1/5, so the perpendicular line will have a slope of -5.

y = -5x + B

-4 = -5(5) + B

B = -4 + 25 = 21

y = -5x + 21

Thepotemich [5.8K]3 years ago

8 0

Answer:

Step-by-step explanation:

You might be interested in

Please help with steps, will give brainliest. Simplify

kumpel [21]

Answer:

=53*(54 ×^9 y^12 z^15)^1/2

=53*y^6 *z^(15/2) * x^( 9/2)* (54)^1/2

4 0

2 years ago

The product of 5 and g

stich3 [128]

Answer:

Step-by-step explanation:

6 0

3 years ago

I need help:/ I’m in college

lesantik [10]

Step-by-step explanation:

Amount of acid = 14.9% of 331 mL solution

= 0.149×(331 mL)

= 49.3 mL acid

3 0

3 years ago

What is the decimal that is equivalent to the expression 3 * 100 + 7 * 10 + 4 * 1 + 6 * 0.01 + 1 * 0.001

barxatty [35]

I believe it is 0.0318

5 0

2 years ago

Find the angle between the vectors u=i-9j and v=8i+5j

Katyanochek1 [597]

Answer: $\cos^{-1} \left(\frac{-37}{\sqrt{7298}} \right) \approx 115.665^{\circ}$

Step-by-step explanation:

$u \cdot v=(1)(8)+(-9)(5)=-37\\\\|u|=\sqrt{1^{2}+(-9)^{2}}=\sqrt{82}\\\\|v|=\sqrt{8^{2}+5^{2}}=\sqrt{89}\\\\\cos \theta=\frac{-37}{\sqrt{82}\sqrt{89}}\\\\\theta=\cos^{-1} \left(\frac{-37}{\sqrt{82}\sqrt{89}} \right)\\\\\theta=\boxed{\cos^{-1} \left(\frac{-37}{\sqrt{7298}} \right) \approx 115.665^{\circ}}$

7 0

2 years ago