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2 years ago

5^2x-4×5x^+1-125=0. Pls how do I solve it very quickly???.

Mathematics

1 answer:

Arada [10]2 years ago

7 0

Answer:

x=25

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

(52)(x)−(4)(5)x1−125=0

25x+−20x+−125=0

(25x+−20x)+(−125)=0(Combine Like Terms)

5x+−125=0

5x−125=0

Step 2: Add 125 to both sides.

5x−125+125=0+125

5x=125

Step 3: Divide both sides by 5.

5x/5=125/5

x=25

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Rose wants to buy a toy that costs $14. If she gives the cashier $20, which equation will tell you how much change Rose gets bac

irga5000 [103]

Answer:

$20-$14=$6

Step-by-step explanation:

this is correct because if you give the $20 to them and the toy costs $14 then you subtract that from the amount you gave them then whatever is left is how much you get back.

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2 years ago

Hannah just put a 36-foot long

VARVARA [1.3K]

Answer:

perimeter=sum of all sides

p=2(L+W)

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8 0

2 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

3 years ago

Ted and Jude are each saving money each month. After x months, the amount of money, in dollars, that Ted has saved is represente

Nadusha1986 [10]

Answer:

True Statements are -

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

Step-by-step explanation:

Given - Ted and Jude are each saving money each month. After x months, the amount of money, in dollars, that Ted has saved is represented by the

equation T = 40x, and the amount of money that Jude has saved is represented by the equation J = 60x.

To find - Choose all of the statements that are true.

A. Each month, Jude saves two-thirds as much money as Ted saves.

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

C. The amount of money Jude saves each month is $20 more than the amount Ted saves each month.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

Proof -

Given that,

After x months,

Ted saved the amount of money, T(x) = 40x

Jude saved the amount of money, J(x) = 60x

Now,

In 1 month,

Ted saved money, T(1) = 40(1) = 40

Jude saved money, J(1) = 60(1) = 60

So,

In 1st month, Jude saved money 20 more than Ted saved.

Now,

In 2 month,

Ted saved money, T(2) = 40(2) = 80

Jude saved money, J(2) = 60(2) = 120

So,

In 2nd month, Jude saved money 40 more than Ted saved.

Now,

In 3 month,

Ted saved money, T(3) = 40(3) = 120

Jude saved money, J(3) = 60(3) = 180

So,

In 3rd month, Jude saved money 60 more than Ted saved.

So,

Option C is incorrect

Because

In 1st month, Jude saves is $20 more than the amount Ted saves

In 2nd month, Jude saves is $40 more than the amount Ted saves

Now,

In 1st month,

Ted saves = 40

and

$\frac{2}{3}(40)$ = 26.67

So, Jude will save = 40 + 26.67 = 66.67

But Jude saves 60

So,

Option A is incorrect.

i.e. A. Each month, Jude saves two-thirds as much money as Ted saves.

Now,

We can see that,

In 3 months, Ted will have saved the money = 120

In 2 months, Jude will have saved the money = 120

So,

Option B is correct.

i.e. B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

Also,

We can see that

In 1st month, Jude saved money 20 more than Ted saved.

In 2nd month, Jude saved money 40 more than Ted saved.

In 3rd month, Jude saved money 60 more than Ted saved.

So,

Option D is correct.

i.e. D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

∴ we get

True Statements are -

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

7 0

2 years ago

Please solve correctly<br> full explanation please<br> will give brainlist

rosijanka [135]

Answer:

They subtracted the values incorrectly

Step-by-step explanation:

The formula for finding the distance between two points is this:
$\sqrt{} (x_{2} -x_{1})^{2} + (y_{2} -y_{1} )^{2}$

The person subtracted the variables incorrectly and subtracted x1 from x2 and y1 from y2 instead of the other way around.

To correctly solve it, do this: $\sqrt{((-7-(-1))^2+(-6-2)^2} = \sqrt{(-6)^{2} + (-8)^{2} } = \sqrt{36+64} = \sqrt{100} = 10$

Hope this helped!

7 0

2 years ago

Read 2 more answers