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2 years ago

9

The product of 5 and a number x is 1/4 What is the value of x?

1 answer:

attashe74 [19]2 years ago

8 0

Answer:

x=19/4

Step-by-step explanation:

The product of 5 and a number x is 1/4

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A sporting goods store buys a basketball for $10.00 and then adds a 110% markup to the price. How much will the basketball sell

sweet-ann [11.9K]

Answer: The basketball will sell for $21 .

Step-by-step explanation:

Hi, to answer this question, first, we have to multiply the cost of the basketball (10) by the percent markup in decimal form (divided by 100).

Mathematically speaking:

10 x (110/100) = $11

Now, we have to add the markup amount (11) to the cost (10) to obtain the selling price:

10 +11 = $21

The basketball will sell for $21 .

3 0

3 years ago

Read 2 more answers

Solve 5(2^x+4)=15 round to the nearest thousandth

Novosadov [1.4K]

Answer:

Step-by-step explanation:

Begin by dividing by 5

2x + 4 = 3

Is this written as 2^(x + 4) = 3? I think it is.

Take the log of both sides

log 2^(x + 4) = log(3)

(x + 4) * log(2) = log 3

log 2 = 0.30103

log 3 = 0.47712

(x + 4) = log2 / log3

x + 4 = 0.63093 Add 4 to both sides

x = -3.369 Rounded to the nearest thousandth

====================

If you mean the question exactly as it is written (the 4 is not part of the power)

5(2^x + 4) = 15

2^x + 4 = 3

2^x = 3 - 4

2^x = - 1

This can't be done 2 to any power should be >0.

if x>0 then this will give an ever increasing number

if xK0 then this will give an ever decreasing answer but still greater than 0.

No value will make 2^x go to something minus.

If I have misread this in some way, leave a note and I will get back to you.

5 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

A can of tuna has a volume 18pi cm3 and a height of 2 cm. Find the area of the label that wraps around the entire can and does n

Sphinxa [80]

Answer:

The area of the label is $12\pi\ cm^{2}$

Step-by-step explanation:

step 1

Find the radius

we know that

The volume of the cylinder (can of tuna) is equal to

$V=\pi r^{2} h$

In this problem we have

$V=18\pi\ cm^{3}$

$h=2\ cm$

substitute and solve for r

$18\pi=\pi r^{2} (2)$

Simplify

$18=r^{2}(2)$

$r^{2}=9$

$r=3\ cm$

step 2

<em>Find the lateral area of the can</em>

The lateral area of the cylinder (can of tuna) is equal to

$LA=2\pi rh$

we have

$r=3\ cm$

$h=2\ cm$

$LA=2\pi (3)(2)=12\pi\ cm^{2}$

5 0

3 years ago

A store bought jeans at a cost of $40 a pair. Select all the statements that are true.

Serhud [2]

Where are the answer options

4 0

3 years ago