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4 years ago

-10y+9x=-9 10y+5x=-5

2 answers:

8 0

Point Form:
(-1,0)
Equation Form:
x=−1, y=0
There weren’t specific instructions to answer this question so here’s what x and y would equal

TEA [102]4 years ago

7 0

You cannot add -10y+9x or 10y+5x because of the rule “combine like terms”...

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Mr. Wonka optimized your team’s box to maximize chocolate bar sales. The new volume is (x^3-9x) inches ^3, the length is (x+3) a

lyudmila [28]

Answer:

$x^2-3x$

Step-by-step explanation:

If we use polynomial division

$x^2-3x$

$x+3$ | $x^3+0x^{2} -9x$

$x^3+3x^2$

$-3x^2-9x$

$-3x^2-9x$

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3 years ago

Mutiple choice (needed 20 characters)

andrezito [222]

The answer is the C=10p

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3 years ago

How many times does 3 go in to 83

jok3333 [9.3K]

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3 years ago

Read 2 more answers

Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T w

Alinara [238K]

Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100× $(\frac{1}{2} )^{3.322}$ .

Step-by-step explanation:

The equation for half life is of the form

A = A₀× $(\frac{1}{2} )^{\frac{t}{h} }$ .........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ $\frac{t}{h}$ and and the equation (2) can be written as

Percentage remaining after 2 half lives is

$\frac{A}{A_0}$ ×100=100× $(\frac{1}{2} )^{2 }$

However if the half life of Technetium-99m is 6 hours then we have for one day

$\frac{A}{A_0}$ ×100=100× $(\frac{1}{2} )^{2 *2}$

Therefore an expression that represents the number of days until only 10% remains is

$\frac{A}{A_0}$ ×100=100× $(\frac{1}{2} )^{\frac{d}{h} }$ = 10 %

$(\frac{1}{2} )^{\frac{d}{h} }$ = $\frac{1}{10}$

= ㏑ $(\frac{1}{2} )^{\frac{d}{h} }$ = ㏑( $\frac{1}{10}$ )

= $\frac{d}{h}$ ×㏑ $(\frac{1}{2} )$ = ㏑( $\frac{1}{10}$ )

$\frac{d}{h}$ = $\frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )}$ = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100× $(\frac{1}{2} )^{3.322}$

3 0

4 years ago

help The screen in a theatre is 22 ft high and is positioned 10 ft above the floor, which is flat. The first row of seats is 7 f

Stella [2.4K]

(a.)
Let's say α is the angle that subtends from the top of the screen to horizontal eye-level.
Let β be the angle that subtends from the bottom of the screen to horizontal eye-level.

tanα = (22 + 10 - 4) / x = 28/x
α = arctan(28/x)

tanβ = (10 - 4) / x = 6/x
β = arctan(6/x)

Ɵ = α - β
Ɵ = arctan(28/x) - arctan(6/x)

(b.)
tanƟ = tan(α - β) = (tanα - tanβ) / (1 + tanα tanβ)
tanƟ = (28/x - 6/x) / [1 + (28/x)(6/x)]
tanƟ = (22/x) / [1 + (168/x²)]
tanƟ = 22x / (x² + 168)
Ɵ = arctan[22x / (x² + 168)]

4 0

3 years ago