All categories

2 years ago

I need help with these two pls

Mathematics

1 answer:

natita [175]2 years ago

3 0

Answer:

1)D

2)A

Step-by-step explanation:

1) $-7a-2x+5a-6x\\-7a+5a-2x-6x\\-2a-8x$

The last choice is the correct answer.

$16bc-8ab=8b(2c-a)$

The answer is the first choice

PLEASE MARK ME AS BRAINLIEST

You might be interested in

Which pair of changes does not involve forming or breaking chemical bonds? Question 4 options: burning and melting rusting and c

umka2103 [35]

Answer:

boiling and melting

Step-by-step explanation:

7 0

3 years ago

For the family picnic, Dana's mom buys a box of 12 forks, a box of 18 spoons, and a box of 24 knives. How many boxes of each wil

SOVA2 [1]

Answer:

Step-by-step explanation:

This is assuming the following:

1) fork boxes comes with 12 count.

2) spoon boxes comes in 18 count.

3) knife boxes comes in 24 count.

Find the least common factor:

12: 12, 24, 36, 48, 60, 72

18: 18, 36, 54, 72

24: 24, 48, 72

72 is the least common factor shared by all these numbers.

3 0

2 years ago

Read 2 more answers

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second