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3 years ago

PLEASE HELP!!!! ASAP grade 9 work

Mathematics

2 answers:

Rom4ik [11]3 years ago

4 0

Answer:

Step-by-step explanation

x . y

-1 . 1

0 . 3

1 . 5

2 . 7

3 9

Licemer1 [7]3 years ago

4 0

Answer:

When x = -1, y = 1.

When x = 0, y = 3.

When x = 1, y = 5.

When x = 2, y = 7.

When x = 3, y = 9.

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Help me out here plssss

Minchanka [31]

Answer:

Step-by-step explanation:

SV = RV

2x+2 = 10

x = 4

r² = RV² + VZ² = 10²+7.5² = 156.25

UZ = √[r²- (QR/2)²] = √[156.25 - 10²] = √56.25 = 7.5

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3 years ago

What os the best definition of a triangle

pav-90 [236]

D. Becaus it is the only one that mentions the characters of a triangle

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3 years ago

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Use a table of values with at least 5 values to graph the following function:

34kurt

The given expression :

$y=(\frac{1}{2})^x$

For coordinates:

put x = 0 then :

$\begin{gathered} y=(\frac{1}{2})^0 \\ y=1 \end{gathered}$

Coordinate : (x, y) = (0, 1)

Put x= 1 and simplify :

$\begin{gathered} y=(\frac{1}{2})^1 \\ y=\frac{1}{2} \\ y=0.5 \end{gathered}$

Coordinate : (x, y) = ( 1, 0.5)

Put x = (-2) and simplify :

$\begin{gathered} y=(\frac{1}{2})^{-2} \\ y=\frac{1^{-2}}{2^{-2}} \\ y=\frac{2^2}{1^2} \\ y=2^2 \\ y=4 \end{gathered}$

Coordinate : (x, y) = ( -2, 4)

Put x = (-3) and simplify :

$\begin{gathered} y=(\frac{1}{2})^{-3} \\ y=\frac{1^{-3}}{2^{-3}} \\ y=\frac{2^3}{1^3} \\ y=2^3 \\ y=8 \end{gathered}$

Coordinate : (x, y) = (-3, 8)

Substitute x = (-1) and simplify :

$\begin{gathered} y=(\frac{1}{2})^x \\ y=(\frac{1}{2})^{-1} \\ y=\frac{1^{-1}}{2^{-1}} \\ y=\frac{2}{1} \\ y=2 \end{gathered}$

Coordinate : (x, y) = ( -1, 2)

So, the coordinates are :

The graph is :

4 0

2 years ago

Evaluate the given integral by changing to polar coordinates. 8xy dA D , where D is the disk with center the origin and radius 9

BabaBlast [244]

Answer:

Step-by-step explanation:

∫∫8xydA

converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.

So,

∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ

So we integrate r from 0 to 9 and θ from 0 to 2π.

∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ

= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ

= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ

= 8[6561/4]∫(cosθsinθ)dθ

= 13122∫(cosθsinθ)dθ

Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2

Substituting this we have

13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ

= 13122/2[-cos2θ]/2 from 0 to 2π

13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]

= -13122/4[cos4π - cos(0)]

= -13122/4[1 - 1]

= -13122/4 × 0

= 0

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3 years ago

Rearrange to make x the subject

Zolol [24]

Answer
t3/b=x
Step by step explanation
t = bx/3
Take b/3 to the left hand side
= x = 3t/b
Hope it helped !!
Comment below

7 0

3 years ago

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PLEASE HELP!!!! ASAP *grade 9 work*

PLEASE HELP!!!! ASAP grade 9 work