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Georgia [21]
3 years ago
12

Which statement about the ordered pairs (2, −9) and (3, −6) is true for the equation 5x−y3=13 ?

Mathematics
2 answers:
Hatshy [7]3 years ago
5 0
C is correct..........
pentagon [3]3 years ago
4 0

Answer:

THe aNSweR Is...

Step-by-step explanation: NeiTHeR orDerEd PaiR iS a sOLutIoN.


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In triangle PQR, find the measure of angle Q.<br> P= 64 degrees<br> Q= X<br> R= 2x
gizmo_the_mogwai [7]

Answer:

38.667°

Step-by-step explanation:

the total sum of 3 angles in a triangle is 180

in triangle PQR this means:

P+Q+R = 180

so:

64+x+2x=180

64+3x=180

3x=116

x=116/3=38.667

So Q=X=38.667

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2 years ago
Without plotting the point (-3, -4), determine in which quadrant
Mandarinka [93]

Answer:

c

Step-by-step explanation:

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Factor the expression<br> 6x^4 - 162x
Vesnalui [34]
Take out common numbers/variables in both terms;

6x(x³-27) 

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3 years ago
Read 2 more answers
The area of a square floor on a scale drawing is 64 square centimeters, and the scale drawing is 1 centimeter:3 ft. What is the
Oxana [17]

Answer:

Part a) The area of the actual floor is 576\ ft^{2}

Part b) The ratio of the area in the drawing to the actual area is \frac{1}{9}\frac{cm^{2}}{ft^{2}}

Step-by-step explanation:

we know that

The scale drawing is \frac{1}{3}\frac{cm}{ft}

step 1

Find the dimensions of the square on a scale drawing

The area of a square is equal to

A=b^{2}

where

b is the length side of the square

A=64\ cm^{2}

so

64=b^{2}

b=8\ cm

step 2

Find the dimensions of the actual floor

Divide the length of the floor on the drawing by the scale drawing

8/(1/3)=24\ ft

step 3

Find the area of the actual floor

The area of a square is equal to

A=b^{2}

substitute

A=24^{2}=576\ ft^{2}

step 4

Find the ratio of the area in the drawing to the actual area

\frac{64}{576}\frac{cm^{2}}{ft^{2}}

Simplify

Divide by 64 both numerator and denominator

\frac{1}{9}\frac{cm^{2}}{ft^{2}}

3 0
3 years ago
Anyone know how to do these? Show work if you do
Charra [1.4K]
4. Is a building
5. Is a pizza
3 0
3 years ago
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