All categories

3 years ago

I need help, algebra is hard

Mathematics

2 answers:

jasenka [17]3 years ago

5 0

Answer:

Step-by-step explanation:

aleksandr82 [10.1K]3 years ago

3 0

Answer: C

Step-by-step explanation:

I think this one is it?

You might be interested in

Rewrite 8^6 • 8^4 in the form 8^n

Ede4ka [16]

Answer:

8^10

Step-by-step explanation:

Keep it as in power form. Note that the base is the same (8).

When the case is that the base is the same:

If multiplying, add the powers together.

If dividing, subtract the powers.

8^6 * 8^4 = 8^(6 + 4) = 8^10

8^10 is your answer.

5 0

3 years ago

Read 2 more answers

Solve for v.<br> -8v+3(v-3) = 16

puteri [66]

Hi there! :)

Answer:

$\huge\boxed{v = -5}$

-8v + 3(v - 3) = 16

Distribute the 3:

-8v + 3(v) + 3(-3) = 16

-8v + 3v - 9 = 16

Combine like terms and simplify:

-5v - 9 = 16

-5v - 9 + 9 = 16 + 9

-5v = 25

Divide both sides by -5:

-5v/(-5) = 25/(-5)

v = -5.

7 0

2 years ago

Read 2 more answers

A shoreline observation post is located on a cliff such that the observer is 280 feet above sea level. The observer spots a ship

morpeh [17]

Answer:

Step-by-step explanation:

a) D/280 = tan(90 - 6)

D = 280tan84 = 2,664 ft

b) D/280 = tan(90 - 16)

D = 280tan74 = 976 ft

ship speed = (2664 - 976) / 43 = 39.2 ft/s

4 0

2 years ago

Find the point where the line crosses the x-axis. y = –5x + 5

Ronch [10]

Every point on the x-axis has y-coordinate 0.
Let y = 0, and solve for x.

y = -5x + 5

0 = -5x + 5

5x = 5

x = 1

The line crosses the x-axis at x = 1.

8 0

2 years ago

What is the distance covered by a train while slowing down from

alukav5142 [94]

Answer:

The distance covered is 113.75 m

Step-by-step explanation:

As per the question:

The initial velocity of the train, v = 20 m/s

The final velocity of the train, v' = 6 m/s

Uniform deceleration, a = 1.6 $m/s^{2}$

Or uniform acceleration, a = - 1.6 $m/s^{2}$

<em>Here, the body decelerates, i.e., slows down at a uniform rate thus we take acceleration with negative sign.</em>

Now, to find the distance covered, s:

Using the eqn of Kinemetics:

$v'^{2} = v^{2} + 2as$

$6^{2} = 20^{2} + 2(- 1.6)s$

$36 - 400 = - 3.6s$

$s = \frac{- 364}{- 3.6} = 113.75\ m$

5 0

3 years ago