All categories

3 years ago

What is the ratio of the two values and what new value do they produce? 4m in 2 min

Mathematics

1 answer:

igomit [66]3 years ago

3 0

The ratio is 2:1. They could be resaigned to 8 meters and 4 meters.

You might be interested in

Could rlly use help ????!!!!!!??

Usimov [2.4K]

Answer: 13

Step-by-step explanation:

180 - (154 + 13) = 13

the triangle is isosceles

7 0

3 years ago

Can anyone help me with this question please ??

Anna007 [38]

Answer:

Try 7986

Step-by-step explanation:

(a+b) × height ÷ 2

(a+b are the parallel sides)

102+140=242 ×66= 15972 ÷2= 7986

7 0

3 years ago

Please help! 2x + 6 (x=4)

Anna007 [38]

Answer:

2(4)+6

8+6

Step-by-step explanation:

7 0

3 years ago

A contractor purchases ceramic tile to remodel a kitchen floor. Each tile cost $4 and the adhesive/grouting material costs $17.8

erma4kov [3.2K]

Answer:

132 tiles

Step-by-step explanation:

Each tile cost $4

The adhesive/grouting material costs $17.82.

If the contractor pays a total of $545.82

Let the number of tiles be represented as x

The equation is represented as:

$17.82 + $4 × x = $545.82

17.82 + 4x = 545.82

Collect like terms

4x = 545.82 - 17.82

4x = 528

Divide bother sides by 4

4x/4 = 528/4

x = 132

The number of ceramic tiles the contractor bought = 132 tiles

8 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago