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Tcecarenko [31]

3 years ago

13

John has 8 coins, all of which are quarters and nickels. In all, they're worth $1.00. How many of each kind of coin does John ha

ve?

1 answer:

Nastasia [14]3 years ago

4 0

Answer:

you will need 3 quarters and 5 nickel

Step-by-step explanation:

3 quarters is 75 and 5 nickels is 25 =1.00

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Solve for x and graph the solution on the number line<br> 15 > 2x – 9 > -23

Monica [59]

The answer is 12>x>-7

Step-by-step explanation:

it's first divided into 2 parts

1 . 15 > 2x - 9

2. 2x - 9 > -23

Then solve them individually.

let's do it,

<u>First half</u>

15 > 2x - 9

15+9 > 2x

24 > 2x

24/2 > x

12>x

<u>Second half</u>

2x-9>-23

2x>-23+9

2x>-14

x>-14/2

x>-7

<u>Joining of the two halves</u>

We have :

12>x>-7

8 0

3 years ago

Is this table proportional or non-proportional?

tatyana61 [14]

Answer:

in my opinon, this table is proportional

5 0

3 years ago

Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

Further analysis of the data for the breakfast cereals in exercise 46 looked for an assocation between fiber content and calorie

photoshop1234 [79]

Complete Question:

Attached below as picture.

Answer:

From first graph there is no linear pattern so here linearity assumption violated.

From second graph there is observation is in some pattern like funnel or v shape so there is no constant variance occur that is there is no constant variance for error.

Constant variance for error occur when in residual plot all observation are in scatter everywhere.

From third graph we can say there is positive distribution but for regression analysis we need symmetric that is normal distribution.

Step-by-step explanation:

See graphs attached below.

8 0

4 years ago

Write in standard form: <img src="https://tex.z-dn.net/?f=%28%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Csqrt%7B2%7D

lutik1710 [3]

Answer:

-1

Step-by-step explanation:

$=(\frac{1}{\sqrt{2}}(1+i))^4\\=\frac{1}{4}(1+i)^2(i+i)^2\\=\frac{1}{4}(2i)(2i)\\=i^2\\=-1$

7 0

3 years ago