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2 years ago

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Which equation could be used to solve the problem? In the summer, Ynez earns $15 per week mowing lawns. At that rate, if she wor

ks 12 weeks, how much will she make (m)?

1 answer:

Hunter-Best [27]2 years ago

7 0

M=15*12. Or, if you want a more general equation for this problem: m=15(w), where w means weeks worked.

You might be interested in

Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.

FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

$\rho = \sqrt{x^{2}+y^{2}+z^{2}}$

$\phi = cos^{-1}\frac{z}{\rho}$

For angle θ:

If x > 0 and y > 0: $\theta = tan^{-1}\frac{y}{x}$ ;

If x < 0: $\theta = \pi + tan^{-1}\frac{y}{x}$ ;
If x > 0 and y < 0: $\theta = 2\pi + tan^{-1}\frac{y}{x}$ ;

Calculating:

a) (4,2,-4)

$\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}$ = 6

$\phi = cos^{-1}(\frac{-4}{6})$

$\phi = cos^{-1}(\frac{-2}{3})$

For θ, choose 1st option:

$\theta = tan^{-1}(\frac{2}{4})$

$\theta = tan^{-1}(\frac{1}{2})$

b) (0,8,15)

$\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}$ = 17

$\phi = cos^{-1}(\frac{15}{17})$

$\theta = tan^{-1}\frac{y}{x}$

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = $\frac{\pi}{2}$

c) (√2,1,1)

$\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}$ = 2

$\phi = cos^{-1}(\frac{1}{2})$

$\phi$ = $\frac{\pi}{3}$

$\theta = tan^{-1}\frac{1}{\sqrt{2} }$

d) (−2√3,−2,3)

$\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}$ = 5

$\phi = cos^{-1}(\frac{3}{5})$

Since x < 0, use 2nd option:

$\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }$

$\theta = \pi + \frac{\pi}{6}$

$\theta = \frac{7\pi}{6}$

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

$r=\sqrt{x^{2}+y^{2}}$

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

$r=\sqrt{4^{2}+2^{2}}$ = $\sqrt{20}$

$\theta = tan^{-1}\frac{1}{2}$

z = -4

b) (0, 8, 15)

$r=\sqrt{0^{2}+8^{2}}$ = 8

$\theta = \frac{\pi}{2}$

z = 15

c) (√2,1,1)

$r=\sqrt{(\sqrt{2} )^{2}+1^{2}}$ = $\sqrt{3}$

$\theta = \frac{\pi}{3}$

z = 1

d) (−2√3,−2,3)

$r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}$ = 4

$\theta = \frac{7\pi}{6}$

z = 3

5 0

3 years ago

A piece of land is 20cm by 5cm. A portion of the land by size 10cm by 2cm was used to cultivate tomatoes. what is the area of th

Sergio039 [100]

Answer:

80cm^2

Step-by-step explanation:

20*5=100

10*2=20

100-20=80

8 0

3 years ago

In the figure, TP and TS are opposite rays. TQ bisects < RTP.

Ket [755]

Answer:

search-icon-image

Class 9

>>Maths

>>Quadrilaterals

>>Quadrilaterals and Their Various Types

>>In Fig. 6.43, if PQ PS, PQ∥ SR, SQR = 2

Question

Bookmark

In Fig. 6.43, if PQ⊥PS,PQ∥SR,∠SQR=28

0

and ∠QRT=65

0

, then find the values of x and y.

463685

expand

Medium

Solution

verified

Verified by Toppr

Given, PQ⊥PS,PQ∥SR,∠SQR=28

∘

,∠QRT=65

∘

According to the question,

x+∠SQR=∠QRT (Alternate angles as QR is transversal.)

⇒x+28

∘

=65

∘

⇒x=37

∘

Also ∠QSR=x

⇒∠QSR=37

∘

Also ∠QRS+∠QRT=180

∘

(Linear pair)

⇒∠QRS+65

∘

=180

∘

⇒∠QRS=115

∘

Now, ∠P+∠Q+∠R+∠S=360

∘

(Sum of the angles in a quadrilateral.)

⇒90

∘

+65

∘

+115

∘

+∠S=360

∘

⇒270

∘

+y+∠QSR=360

∘

⇒270

∘

+y+37

∘

=360

∘

⇒307

∘

+y=360

∘

⇒y=53

∘

Step-by-step explanation:

please mark me as brainlist please

6 0

2 years ago

PLZZZ HELP ME WHAT DOES x=?????? WHOEVER ANSWERS FORST GETS BRAINLIEST THANKYOUUU

Vinil7 [7]

I think it’s 17 cause 51/3 is 17.

6 0

2 years ago

Read 2 more answers

Please help, thanks! Ok so, you pick a card at random, put it back, and then pick another card at random. There are FIVE cards a

tatyana61 [14]

Answer:

$P(x > 1\ and\ x < 2) = \frac{4}{25}$

Step-by-step explanation:

Given

$S = \{1,2,3,4,5\}$

$n(S) = 5$

Required

$P(x > 1\ and\ x < 2)$

$P(x > 1\ and\ x < 2)$ is calculated as:

$P(x > 1\ and\ x < 2) = P(x > 1) * P(x < 2)$

Since it is a probability with replacement, we have:

$P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}$

For x > 1, we have:

$x > 1 = \{2,3,4,5\}\\$

$n(x > 1) = 4$

For x < 2, we have:

$x < 2 = \{1\}$

$n(x < 2) = 1$

$P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}$

becomes

$P(x > 1\ and\ x < 2) = \frac{4}{5} * \frac{1}{5}$

$P(x > 1\ and\ x < 2) = \frac{4}{25}$

6 0

2 years ago