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Leto [7]
3 years ago
6

Need help fast, please

Mathematics
1 answer:
siniylev [52]3 years ago
6 0
Answer: -9x+3y=27
As you can see when you plug in (2,_) into x in both equations -9x+3y=27 comes out with y being a higher number compared to the other equation. Thus the equation -9x+3y=27 is the answer having a higher rate of change!

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Point Z has the coordinates (−3,5) and is translated left 2 units and down 2 units to create point Z′.
iris [78.8K]

Answer:

(-5, 3)

Step-by-step explanation:

it translated left 2 units => x'= x-2

down 2 units=> y'= y-2

8 0
3 years ago
The dot plot below shows the number of wings contestants in a wing-eating contest ate.
Stolb23 [73]

A.18 is the absolute deviation

6 0
4 years ago
Read 2 more answers
Mary works in a factory and earns Rs.955 per month she saves Rs.185 per month. Find the ratio of
Kryger [21]

Answer:

A) 37:191

B)191:154

Step-by-step explanation:\

A) Mary's Savings= 185 Rs

Mary's Income= 955 Rs

Ratio = 185:955

= 185/955

=37/191                     ( INTO LOWEST TERMS )

=37:191

B) Mary's Income= 955 Rs

Mary's Expenditure= ?? = Income- Saving = 955-185= 770

Ratio= 955:770

= 955/770

= 191/154                  ( INTO LOWEST TERMS )

= 191:154

8 0
4 years ago
A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra
Harman [31]

Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em />cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

Step-by-step explanation:

Given

P\ (a,b)

r = \± \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

r = \sqrt{(a)^2 + (b)^2}

Since a belongs to the x axis and b belongs to the y axis;

cos\theta is calculated as thus

cos\theta = \frac{a}{r}

Substitute r = \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}

cos\theta = \frac{a}{\sqrt{a^2 + b^2}}

Rationalize the denominator

cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}

cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

3 0
3 years ago
SOMEONE PLZZZ HELP ME THIS IS DUE IN LIKE 5 MINS (I WILL MARK BRAINLIEST)
andrew-mc [135]

Answer:

1) x = 8

2) ∠RPS = 36°

Step-by-step explanation:

<u>GIVEN :-</u>

  • ∠QPS = 180°
  • ∠QPR = 7x + 88
  • ∠RPS = 3x + 12

<u>TO FIND :-</u>

  1. Value of x
  2. Measure of ∠RPS

<u>FACTS TO KNOW BEFORE SOLVING :-</u>

In a straight line , if there are two angles such that their sum is equal to straight angle (or 180° in other words) , then those angles are known as linear pair.

<u>PROCEDURE :-</u>

1)

Measure of ∠QPS = 180° and it comprises of ∠QPR & ∠RPS.

⇒ ∠QPR & ∠RPS are linear pair.

⇒ ∠QPR + ∠RPS = 180°

⇒ 7x + 88 + 3x + 12 = 180°

⇒ 10x + 100 = 180

⇒ 10x = 180 - 100 = 80

⇒ x = 80/10 = 8

2)

x = 8. So,

∠RPS = 3×8 + 12 = 24 + 12 = 36°

8 0
3 years ago
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