Please help each equation is a number from 1 to 6

miles has 25 coins In his pocket. Some are Nichols and some are dimes. He has a total of $1.65 how many of each type of coin doe

Leni [432]

▪ Answer:

n = 17

d = 8

▪ Step-by-step explanation:

Hi there !

1 nickel = 5c

1 dime = 10c

$1.65 = 165c

n + d = 25 => n = 25 - d

5n + 10d = 165

replace n

5(25 - d) + 10d = 165

125 - 5d + 10d = 165

5d = 165 - 125

5d = 40

d = 8 (dimes)

n = 25 - d

n = 25 - 8

n = 17 (nickels)

Good luck !

7 0

3 years ago

red and blue ceramic tiles are laid according to a pattern such that every 21 red tiles used, 4 blue tiles are used. if a total

ladessa [460]

Answer:

X= blue tiles

Y = red tiles

X/y = 4/21

X+y= 425

Y= 357 tiles

X= 68 tiles

Step-by-step explanation:

red and blue ceramic tiles are laid according to a pattern.

The pattern= every 21 red tiles used, 4 blue tiles are used

Total number of tiles used = 425 tiles

X= blue tiles

Y = red tiles

X/y = 4/21

X+y= 425

21x= 4y

X= 4y/21

X+y= 425

4y/21 + y = 425

25y= 425(21)

Y=( 425*21)/25

Y= 357 tiles

X= 4y/21

X= (4*357)/21

X= 1428/21

X= 68 tiles

4 0

3 years ago

Please help!! What Find the Error and explain why it is wrong!

Nata [24]

Answer:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

Step-by-step explanation:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. The real numerator in Step E should be:

$3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x$

Hence, Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

And further steps are described below:

Step D

$x^{2}-x+4 = 0$

Which according to the Quadratic Formula, represents a polynomial with complex roots. That is: ( $a = 1$ , $b = -1$ , $c = 4$ )

$D = b^{2}-4\cdot a\cdot c$

$D = (-1)-4\cdot (1)\cdot (4)$

$D = -17$ (Conjugated complex roots)

Step E

$(x-0.5-i\,1.936)\cdot (x-0.5+i\,1.936) = 0$

Step F

$x = 0.5+i\,1.936\,\lor\,x = 0.5-i\,1.936$

8 0

3 years ago

Can you help me with these 2 questions?

Alenkinab [10]

Answer:

What school you go to

Step-by-step explanation:

step by step go catch friend and step by step friends

6 0

3 years ago

2+cotA=1 homework help

Alex Ar [27]

$2+\cot A=1\implies \cot A=-1\implies \tan A=-1$

This happens whenever $A=\dfrac{3\pi}4$ or $A=\dfrac{7\pi}4$ . More generally, $\tan A=-1$ whenever you start with one of these angles and add any multiple of $\pi$ , so the general solution would be $A=\dfrac{3\pi}4+n\pi$ , where $n$ is any integer. (Notice that when $n=1$ , you end up with $\dfrac{7\pi}4$ .)

3 0

4 years ago