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2 years ago

R varies directly as s. Find r when s=4 and k=2

1 answer:

5 0

$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
\\
&& y={{ k }}x
\end{array}\\\\
-------------------------------\\\\
thus\implies r=ks\qquad 
\begin{cases}
k=2\\
\uparrow \\
constant\ of\\
variation\\
s=4
\end{cases}\implies r=2\cdot 4$

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Function 1 is defined by the equation y=2x+10

cricket20 [7]

Answer:

C. The functions have the same y-intercept

Step-by-step explanation:

In slope intercept form, y = mx + b, b represents the y-intercept, and in the first function the y-intercept is 10.

The y-intercept is when x = 0, and in the chart, when x equals 0, y equals 10.

10 = 10, so they have the same y-intercept.

5 0

2 years ago

Mr. Jones wants to place a chimney on his roof. It is recommended that the chimney be set at a height of at least 30 feet. The h

Tom [10]

Absolute value of any number is always positive because it represent distance and distance will always positive.

To place chimney at least 30 feet , the distance along the width of the roof (x) will be x = 17.5 feet or x = 2.5 feet.

Since, The height of the roof is described by the function f(x)=−4/3|x−10| + 40

Since, minimum height is 30 feet.

So , substitute f(x)=30 in above expression.

30=−4/3|x−10| + 40

4/3|x−10| = 10

|x−10|=30/4

|x−10| = 7.5

Taking positive case,

x−10 = 7.5

x = 17.5 feet

Taking negative case,

x−10 = - 7.5

x = 10 - 7.5

x = 2.5 feet

Learn more:

brainly.com/question/14473339

7 0

1 year ago

Look at this cone:<br><br>,,

Assoli18 [71]

Slant height=l=3ft
Radius=r=2ft

We know

$\boxed{\sf \star TSA_{(Cone)}=\pi r(r+\ell)}$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=2 \dfrac{22}{7}\times 2(2+3)$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{44}{7}(5)$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{220}{7}$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=31.4ft^2$

Now

New slant height =2(3)=6cm
New radius=2(2)=4cm

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{22}{7}\times 4(4+6)$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{88}{7}(10)$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{880}{7}$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=125.7cm^2$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{125.7}{31.4}$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{4}{1}$

$\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cone)}}:{TSA_{(Old\:Cone)}}=4:1}}}$

7 0

2 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

3 years ago

PLS ANSWER!!! Solve (−5) ⋅ (−3).

BartSMP [9]

Answer:

pretty sure its 15

Step-by-step explanation:

a - by a - = a postive sou yea um

-3X-5 = 15

3 0

2 years ago

Read 2 more answers