All categories

3 years ago

I got the wrong answer, does anyone know what the right answer is?

Mathematics

1 answer:

Tom [10]3 years ago

5 0

It should be C.

The slope formula using 2 point is

m = (y2 - y1) / (x2 - x1)

That's another way of saying C. The problem with B is that it is the upside down of what the slope actually is.

You might be interested in

the equation of line t is 6x+5y=3, and the equation of line q is 5x-6y=0. Which statement about the two lines is true?

nekit [7.7K]

The correct answer would be: 6x+5y=3. I am not entirely sure, but that's what I believe is the correct answer.-Hope this helped!

4 0

2 years ago

Nikhil gets paid a 5 percent commission on every pair of shoes that he sells. He earned $1.00 on the last pair of shoes that he

atroni [7]

Answer: C.20.00

Step-by-step explanation:

I just took the test

6 0

2 years ago

Fred bought 4 liters of liquid laundry detergent 3,250 milliliters of fabric softener

Harrizon [31]

3.250 liters of softener because you have to jump three times according to the ladder method.

8 0

2 years ago

Read 2 more answers

A rocket is launched at 85 ft./s from a launch pad that’s 28 feet above the ground. which equation can be used to determine the

3241004551 [841]

Answer:

$h(t)=-16t^2+85t+28$ is equation of height of rocket.

Option D is correct.

Step-by-step explanation:

Given: A rocket is launched with speed 85 ft/s from a height 28 feet.

Launching a rocket follows the path of parabola. The equation of rocket should be parabolic.

Parabolic equation of rocket is

Formula: $h(t)=\dfrac{1}2gt^2+v_0t+h_0$

g ⇒ acceleration due to gravity (-32 ft/s)

v ⇒ Initial velocity ( $v_0=85\ ft/s$ )

h ⇒ Initial height ( $h_0=28\ feet$ )

h(t) ⇒ function of height at any time t

Substitute the given values into formula

$h(t)=\frac{1}{2}(-32)t^2+(85)t+28$

$h(t)=-16t^2+85t+28$

D is correct.

6 0

3 years ago

Read 2 more answers

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

8 0

2 years ago