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3 years ago

5

Round to the nearest thousands 3475.6993

1 answer:

Andreyy893 years ago

4 0

3475.699

°°°°°°°°°°°°°°°

Am I correct?

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insens350 [35]

Answer:

$- 4( - 4x - 5) = - 4 \times - 4x - ( - 4) \times ( - 5) \\ = 16x - 20$

3 0

2 years ago

Read 2 more answers

Which statement is not true about the pattern shown 2/3 , 4/6, 8/12, 16/24 , ...

natta225 [31]

They are multiplying by two .

8 0

3 years ago

Solve the following word problems.

aleksandr82 [10.1K]

Answer:

Step-by-step explanation:

Let a , b , c & d be the no. of chairs in hall A , B , C & D respectively.

Given,

b = 4a

c = a + 15

d = a + 8 + 4a = 5a + 8

Now,

Total no. of chairs (T) = a + b + c + d

= a + 4a + a + 15 + 5a + 8

= 11a + 23

Now,

Average chairs in each hall (Avg) = T/4

= (11a + 23)/4

7 0

3 years ago

8 to 11 do it with equations thx

Alenkasestr [34]

Answer:

Step-by-step explanation:

8. $4\geq x+\frac{3}{4}$

$4(4\geq x+\frac{3}{4} )$

$16\geq 4x + 3$

$-3$ $-3$

$13\geq 4x$

$\frac{13}{4}\geq \frac{4x}{4}$

$\frac{13}{4} \geq x = x\leq \frac{13}{4}$

$x\leq \frac{13}{4}=3.25$

9. $x-3\leq -8$

$+3$ $+3$

$x\leq -5}$

10. $4\leq x-7$

$+7$ $+7$

$11\leq x=x\geq 11$

11. 6.9 < x - 2.3

+ 2.3 + 2.3

9.2 < x = x > 9.2

Hope this helps!

5 0

2 years ago

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

Read more about related rates at:

brainly.com/question/16981791

8 0

2 years ago