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3 years ago

Ayúdenme porfavor necesito esto para mañana resuelto por el Sistema de Ecuaciones Lineales en

Mathematics

1 answer:

AlekseyPX3 years ago

7 0

Deberias mirar videos en linea

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A submarine traveling 200 meters below the surface of the ocean increases its depth by 45 meters. Adam says that the new locatio

melisa1 [442]

Adam made an error of 400 meters

Explanation:

Since, submarine traveling 200 meters below the surface of the ocean

and when the depth increases by 45 meters then submarine new location is

$200+45=245$ meters below from the surface of ocean.

But Adam says that new location is at -155 meters.

So, error is, $245-(-155)=400$ meters

Learn more:

https://brainly.in/question/13000874

8 0

2 years ago

a beverage is made by mixing 3 parts of water with 5 parts of fruit juice how many parts of water are mixed with 1 part on fruit

Fittoniya [83]

Answer:

$\frac{3}{5}=0.6$

Step-by-step explanation:

Given the initial ratio of 3 parts water to 5 parts fruit juice, you can set up a proportion (equivalent ratios) to determine the amount of water needed with just 1 part of fruit juice:

$\frac{water}{fruit}=\frac{3}{5}=\frac{x}{1}$

cross multiply: 5x = 3

divide: 5x/5 = 3/5 or x = 3/5

3/5 = 0.6 parts water

7 0

3 years ago

Read 2 more answers

Does anyone understand this?. I also have to show my work.

snow_lady [41]

Answer:

Ans A). The graph is shown.

Ans B). 18.3333 C temperature when F is 65 temperature

Ans C). 32 F when the line crosses the horizontal axis

Ans D). Slope of line C= $\frac{5}{9}(F -32)$ is $\frac{5}{9}$

Step-by-step explanation:

Given equation is C= $\frac{5}{9}(F -32)$

Ans A).

For the table,

Take the four value of F as 32,41,50,59.

For F = 32.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(32 -32)$

C=0.

For F = 41.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(41 -32)$

C=05

For F = 50.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(50 -32)$

C=10

For F = 59.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(32 -32)$

C=15

Note: The figure shows a graph of given equation with points.

Ans B). Estimate temperature in C when the temperature in F is 65

For F = 65.

The value of C is

C= $\frac{5}{9}(F -32)$

C= $\frac{5}{9}(32 -32)$

C=18.333333.

Ans C). At what temperature, graph lien cross the horizontal axis

When the line crosses the horizontal axis, C=0

Therefore,

C= $\frac{5}{9}(F -32)$

0= $\frac{5}{9}(F -32)$

0= $(F -32)$

F=32 Temperature.

Ans D). Slope of the line C= $\frac{5}{9}(F -32)$

The slope of line is given by s= $\frac{Y1-Y2}{X1-X2}$

Take points from the table of answer A.

let (32,0) and (41,5) using for slope.

s= $\frac{Y1-Y2}{X1-X2}$

s= $\frac{0-5}{32-41}$

s= $\frac{5}{9}$

Slope of line C= $\frac{5}{9}(F -32)$ is $\frac{5}{9}$

7 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago

Given: F(x) = 2x - 1; G(x) = 3x + 2; H(x) = x 2 Find F{G[H(2)]}

Tamiku [17]

H(2)=x²=2²=4
G[H(2)]=G(4)=3(4)+2=12+2=14
F[G[H(2)]]=F(14)=2(14)-1=28-1=27

Answer: F[G[H(2)]]=27

4 0

3 years ago

Read 2 more answers