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3 years ago

7

What is the relationship between the 6s in 660,472

1 answer:

andreev551 [17]3 years ago

8 0

The first 6 in the number is 10 times the 2nd 6.

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What is the result of adding these two equations 8x-4y=-4 -3x+4y=5

Sergeeva-Olga [200]

Answer:

8x-4y=-4

-3x+4y=5

+

-----------------(adding equation)

5x=9

is a required answer

6 0

3 years ago

Which inequality will have a dashed boundary line with a shaded area above its graph? 7x − y ≤ 3 x − y ≤ 4 2x − 9y > 2 4x + 7

alexdok [17]

Above the graph

look at them

7x-y <span><</span>3
x-y<span><</span>4
2x-9y>2
4x+7y>3

dashed boundary line means it does't include htat means taat the symbol is < or >

cross out first 2

one way is to graph the lines as equations and shade accordingly

2x-9y>2
y<2/9x+2/9
shade on bottom

4x+7y>3
y>-7/4x+3/9
shade on top

aswer is 4x+7y>3

5 0

4 years ago

Giovanni is at the market, he notices that 7 apples cost $4.55. How much does he pay for 9 apples?

geniusboy [140]

You divide $4.55 by 7.
$0.65 times 9 = $5.85

5 0

3 years ago

Read 2 more answers

Small sample: During an economic downturn, companies were sampled and asked whether they were planning to increase their workfor

leva [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The confidence level interval is $0.016 \le C \le 0.404$

Step-by-step explanation:

The sample size is $n = 20$

The number planning to increase workforce is $x = 3$

The confidence level is $c = 98$ %

The value of proportion for a plus 4 method is

$p = \frac{x+2}{n+4}$

substituting values

$p = \frac{3+2}{20+4}$

$p =0.21$

The z-critical value at confidence level of 98% is

$z_{c}=z_{0.98} = 2.33$

This values is obtained from the standard normal table

The confidence level interval can be mathematically represented as

$C =p \ \pm z_{c} * \sqrt{\frac{p(1-p)}{n+4} }$

substituting values

$C = 0.21 \pm 2.33 * \sqrt{\frac{0.21(1- 0.21)}{20 +4} }$

$C = 0.21 \pm 0.194$

=> $0.016 \le C \le 0.404$

7 0

3 years ago

Please help me answer this question

stealth61 [152]

The value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

<h3>Differentiation</h3>

From the question, we are to determine dy/dx for the given functions

(i) $x^{2} sin^{2}2x$

Let $u = x^{2}$

and

$v = sin^{2} 2x$

Also,

Let $w= sin2x$

∴ $v = w^{2}$

First, we will determine $\frac{dv}{dx}$

Using the Chain rule
$\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}$

$v = w^{2}$

∴ $\frac{dv}{dw} =2w$

Also,

$w= sin2x$

∴ $\frac{dw}{dx} =2cos2x$

Thus,

$\frac{dv}{dx} = 2w \times 2cos2x$

$\frac{dv}{dx} = 2sin2x \times 2cos2x$

$\frac{dv}{dx} = 4sin2x . cos2x$

Now, using the product rule

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

From above

$u = x^{2}$

∴ $\frac{du}{dx}=2x$

Now,

$\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)$

$\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $xy^{2}+y^{2}x^{3} +2=0$

Then,

$x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0$

$2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0$

$2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}$

$\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }$

$\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Hence, the value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Learn more on Differentiation here: brainly.com/question/24024883

#SPJ1

8 0

2 years ago