All categories

2 years ago

Which of the following are like terms?

Mathematics

2 answers:

vladimir2022 [97]2 years ago

8 0

The 1, 2, 3 I’m not sure bout the 4th one

kolezko [41]2 years ago

3 0

Dde dwsadqddwwxqwrffrfrfrg4d

You might be interested in

Cece is finding the surface area of a square pyramid by finding the sum of the areas of its faces. Which statement best describe

trapecia [35]

Based on the options given, the most likely answer to this query is one square and four triangles. The square triangle has one square which is at the bottom and three triangles above.

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

8 0

2 years ago

Read 2 more answers

The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that

ICE Princess25 [194]

Answer:

The probability that event B will occur is 0.45

Step-by-step explanation:

Given;

probability that event A occurs, P(A) = 0.4

the probability that events A and B both occur, P(A ∩ B) = 0.25

the probability that either event A or event B occurs, P(A ∪ B) = 0.6

To determine the probability that event B will occur, we use probability addition rule;

P(A) + P(B) = P(A ∩ B) + P(A ∪ B)

0.4 + P(B) = 0.25 + 0.6

0.4 + P(B) = 0.85

P(B) = 0.85 - 0.4

P(B) = 0.45

Therefore, the probability that event B will occur is 0.45

4 0

2 years ago

PLEASE HELP I WILL GIVE THE BRAINLIEST

Dmitry [639]

Kdicneiddhd put more points worie dodveidhew

8 0

2 years ago

Read 2 more answers

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$