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2 years ago

14-6=8 what is the tens factor used 10- what = what

Mathematics

1 answer:

Mariulka [41]2 years ago

8 0

Speak Engrish son lol what=what

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Does anyone understand how to solve this?

Nataliya [291]

a f(a) is the function f(x) where x is replaced by a.

So we have

(3x^2 + x + 3 - (3a^2 + a + 3)) / ( x - a)

= (3x^2 - 3a^2 + x - a) / (x = a) Answer

b (3(x + h)^2 + x + h + 3 - (3x^2 + x + 3)) / h

= (3x^2 + 6xh + 3h^2 + x + h - 3x^2 - x - 3) ) / h

= (6xh + h + 3 h^2) / h

= 6x + 3h + 1 Answer

6 0

2 years ago

5 POINTS !! ANSWER FASTTT

MariettaO [177]

N= 2

Substitute 2 in for n in expression

=(5 + n) ÷ (n - 1)
=(5 + 2) ÷ (2 - 1)
=7 ÷ 1
=7

If n= 2, the expression equals 7.

Hope this helps! :)

3 0

2 years ago

In a triangle, the largest angle is always opposite the (Blank) and the smallest angle is always opposite the (Blank) .

Shtirlitz [24]

Answer:

the largest angle is opposite the largest side and the shortest angle is opposite the shortest side

Step-by-step explanation:

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2 years ago

The senior class sells movie tickets to students and teachers. The student

Tasya [4]

D..............................

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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