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3 years ago

Assume that A varies directly with z. If A = 30 when z = 5, find A when z = 9.

2 answers:

6 0

If A varies directly with z, and A is 30 and z is 5, then when z is 1, A is 6.
z: 5 ÷ 5 = 1
A: 30 ÷ 5 = 6
Then, when z is 1, A is 6
So, when z is 9, A is...
z: 1 × 9 = 9
A: 6 × 9 = 54

A is 54

adell [148]3 years ago

5 0

We can this two ways:

1) If A = 30 when Z = 5, then A = 6Z or A = 6*z

So if Z = 9, 9 * 6 = 54. A = 54.

2) You can set up a proportion:
$\frac{30}{5} = \frac{x}{9}$ where x = the unknown A-value.

Crossmultiply:
5x = 30*9
5x = 270

Divide by 5 on both sides to isolate the variable
x = 54

54 is your A value when Z = 9.

Either way you solve it, your answer is 54.

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The perimeter of a rectangular construction site is 124 meters. The width is eight meters more than 5 times the length. Find the

tresset_1 [31]

Answer:

The perimeter of a rectangle is the sum of both lengths and both widths, which is equal to 54 meters. Let's call Length L and Width W.

The question is saying this: L = 3 meters + 3(W). We have 2 variables, which means we need at least 2 equations to solve. So far we have one, our second equation is from the perimeter.

2 lengths + 2 Widths = 54. Now, it's just a plug and chug.

2(3 + 3W) + 2W = 54.

6 + 6W + 2W = 54

8W = 48

W=6

L = 3 + 3(6) = 21

To double check: 2(21) + 2(6) = 42 + 12 = 54

The Width is 6 meters, and the Length is 21 meters.

7 0

2 years ago

PLEASE HELP!! Name the solution to this system of equations:

RoseWind [281]

9514 1404 393

Answer:

(x, y) = (3, -1)

Step-by-step explanation:

The solution to the system of equations is the point where the lines intersect.

(x, y) = (3, -1)

3 0

3 years ago

5-2x=15 <img src="https://tex.z-dn.net/?f=5%20-%202x%20%3D%2015" id="TexFormula1" title="5 - 2x = 15" alt="5 - 2x = 15" align

padilas [110]

5-2x=15

 $5-2x=15 \\ \\ -2x= 15-5 \\ \\ -2x= 10 \\ \\ x= \dfrac{10}{-2} \to \boxed{x= -5}$

6 0

3 years ago

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$