If it takes

PLEASE HELP, The blueprints for the johnsons living room has a scale of 2 inches= 5 feet. The family wants to use a scale of 1 i

olchik [2.2K]

Because we got five feet for two inches at preceding case that mean we had 2.5 feet for each inch by dividing 5 on 2 so that will give us the result 2.5 standing as well as that scale change from 2.5 to 3 for each inch in the new blueprint so by multiplying 3 feet by 2 inch that give us 6 feet width

7 0

3 years ago

(6x-5y+4)dy+(y-2x-1)dx=0

Len [333]

(6x - 5y + 4) dy + (y - 2x - 1) dx = 0

(6x - 5y + 4) dy = (2x - y + 1) dx

dy/dx = (2x - y + 1) / (6x - 5y + 4)

Let X = x - a and Y = y - b. We want to find constants a and b such that

dY/dX = (a rational function)

where the numerator and denominator on the right side are free of constant terms. Substituting x and y in the equation, we have

dY/dX = (2 (X + a) - (Y + b) + 1) / (6 (X + a) - 5 (Y + b) + 4)

dY/dX = (2X - Y + 2a - b + 1) / (6X - 5Y + 6a - 5b + 4)

Then we solve for a and b in the system,

2a - b + 1 = 0

6a - 5b + 4 = 0

==> a = -1/4 and b = 1/2

With these constants, the equation reduces to

dY/dX = (2X - Y) / (6X - 5Y)

Now substitute Y = VX and dY/dX = X dV/dX + V :

X dV/dX + V = (2X - VX) / (6X - 5VX)

The equation becomes separable after some simplification:

X dV/dX + V = (2 - V) / (6 - 5V)

X dV/dX = (2 - V) / (6 - 5V) - V

X dV/dX = (2 - V - (6 - 5V)) / (6 - 5V)

X dV/dX = (4V - 4) / (6 - 5V)

- (5V - 6) / (4V - 4) dV = 1/X dX

Integrate both sides:

-5/4 V + 1/4 ln|4V - 4| = ln|X| + C

Extract a constant from the logarithm on the left:

-5/4 V + 1/4 (ln(4) + ln|V - 1|) = ln|X| + C

-5/4 V + 1/4 ln|V - 1| = ln|X| + C

-5V + ln|V - 1| = 4 ln|X| + C

Get this back in terms of Y :

-5Y/X + ln|Y/X - 1| = 4 ln|X| + C

Now get the solution in terms of y and x :

-5 (y - 1/2)/(x + 1/4) + ln|(y - 1/2)/(x + 1/4) - 1| = 4 ln|x + 1/4| + C

With some manipulation of constants and logarithms, and a bit of algebra, we can rewrite this solution as

-5 (4y - 2)/(4x + 1) + ln|(4y - 4x - 3)/(4x + 1)| = 4 ln|x + 1/4| + 4 ln(4) + C

-5 (4y - 2)/(4x + 1) + ln|(4y - 4x - 3)/(4x + 1)| = 4 ln|4x + 1| + C

-5 (4y - 2)/(4x + 1) + ln|4y - 4x - 3| - ln|4x + 1| = 4 ln|4x + 1| + C

-5 (4y - 2)/(4x + 1) + ln|4y - 4x - 3| = 5 ln|4x + 1| + C

8 0

3 years ago

425 is what percent of 1700

IgorLugansk [536]

Find 425/1700 as a decimal fraction, and then mult. the result by 100%:

425
-------- * 100% = 25% (answer)
1700

4 0

4 years ago

Verify the trigonometric identities

snow_lady [41]

1)

here, we do the left-hand-side

$\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2
\\\\\\\
[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]
\\\\\\
2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2$

2)

here we also do the left-hand-side

$\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)
\\\\\\
\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}
\\\\\\
\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)$

3)

here, we do the right-hand-side

$\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}
\\\\\\
\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}
\\\\\\
\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}$

8 0

3 years ago

I need a answer A.S.A.P PLEASE WILL GIVE 19 POINTS

Ipatiy [6.2K]

Answer:

Okie when you do these you got to give all information

Step-by-step explanation: because people can help you instead of looking at a picture wonderin what they are they are supposed to do

3 0

3 years ago