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3 years ago

What is the distance between the points?! round to the nearest tenth i need to show work

Mathematics

1 answer:

Contact [7]3 years ago

5 0

Answer:

8.6

Step-by-step explanation:

To find the distance between two points we use the formula posted below

All we need to do is figure out what the points are on the graph and plug them into the formula... we end up with

the square root of (5-(-2)^2+(2-(-3)^2 and get the answer of 8.602325267

then we round to the nearest tenth and get 8.6

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there are 98 girls at Ross elementary there are 15 less boys then girls how many boys and girls are at Ross elementary

Luba_88 [7]

Answer: 98-15= 83 boys

8 0

3 years ago

If y = 3x^3 - 2x and dx/dt equals 3, find dy/dt when x = -2. Give only the numerical answer. For example, if dy, dt = 4, type on

Nadya [2.5K]

Answer:

$\frac{dy}{dt} = 102$

Step-by-step explanation:

step(i)

Given function y = 3 x³ - 2 x ... (i)

Differentiating equation (i) with respective to 'x'

$\frac{dy}{dx} = 3 (3 x^{2} ) - 2(1)$

Given x = -2

$(\frac{dy}{dx} ) x_{=-2} = 3 (3 (-2)^{2} ) - 2(1)$

$(\frac{dy}{dx} ) x_{=-2} = 36 -2 =34$

Step(ii):-

we know that

$\frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

Given $\frac{dx}{dt} = 3 and \frac{dy}{dx} = 34$

$\frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

substitute values $\frac{dx}{dt} = 3 and \frac{dy}{dx} = 34$

⇒ $34 = \frac{\frac{dy}{dt} }{3 }$

cross multiplication , we get

$\frac{dy}{dt} = 34( 3 ) = 102$

Final answer:-

$\frac{dy}{dt} = 34( 3 ) = 102$

8 0

3 years ago

Solve for X.........

Morgarella [4.7K]

Answer:

Hey there!

There is this equation, which states that the angle created by an inscribed angle is half the measure of the arc.

Thus, we can write: 60=5x-5

65=5x

x=13

Let me know if this helps :)

5 0

3 years ago

Read 2 more answers

Find the value of x that makes lines u and v parallel.

omeli [17]

Answer:

x = 5

Step-by-step explanation:

From the picture attached,

If the lines u and v are parallel and a third line (transversal) is intersecting these lines at two distinct points,

Both the angles given in the picture, will be equal in measure as these angles are interior alternate angles.

16x = 15x + 5

16x - 15x = 5

x = 5

Therefore, x = 5 makes the lines v and u parallel.

3 0

3 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

4 years ago