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2 years ago

1/2(10x+20y+10z help me please

Mathematics

2 answers:

den301095 [7]2 years ago

6 0

Given is 1/2(10x+20y+10z)

Simplifying the given expression using Algebra properties.

$\frac{1}{2} (10x+20y+10z) \\\\Step \;1: \;Using \;Distribution \;property \\\\=\frac{10x}{2}+ \frac{20y}{2}+ \frac{10z}{2} \\\\Step \;2: \;Dividing \\\\=5x+10y+5z \\\\Step \;3: \;Factoring \;out \;common \;factor \\\\= 5*(x+2y+z)$

Hence, the final answer would be (5x + 10y + 5z) or 5·(x + 2y + z).

kupik [55]2 years ago

4 0

Answer:

(5x+10y+5z)

Step-by-step explanation:

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2 years ago

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madreJ [45]

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3 years ago

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sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

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3 years ago

Felicia wants to build a kite with the shape shown. If AC is 60 cm, how many centimeters are in the length of BD?

Kay [80]

Answer:

Step-by-step explanation:

By applying tangent rule in the given right triangle AOB,

tan(30°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

$\frac{1}{\sqrt{3}}=\frac{BO}{OA}$

$OA=BO(\sqrt{3})$

By applying tangent rule in the given right triangle BOC,

tan(60°) = $\frac{OC}{BO}$

OC = BO(√3)

OA + OC = AC

$BO(\sqrt{3})+BO(\sqrt{3}) =60$

2√3(BO) = 60

BO = 10√3

OC = BO(√3)

OC = (10√3)(√3)

OC = 30

By applying tangent rule in right triangle DOC,

tan(60°) = $\frac{OD}{OC}$

OD = OC(√3)

OD = 30√3

Since, BD = BO + OD

BD = 10√3 + 30√3

BD = 40√3

≈ 69.3

6 0

2 years ago

491,852 nearest hundred thousand

r-ruslan [8.4K]

500,000

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3 years ago