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2 years ago

Help please, i need a maths wizz!! :( (See pinned pics for questionS)

Mathematics

1 answer:

lana66690 [7]2 years ago

8 0

Answer:

200 people attended screen 1

150 to screen 2 and 150 to screen 3

for question 2:it's 265.1

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A population of rabbits is described by the function R(t) = 100(2t/5), where t is measured in months and R is measured in rabbit

Savatey [412]

Answer and Step-by-step explanation: The graph is shown in the attachment.

a. ΔR on [1,2] is mathematically expressed as:

ΔR = R(2) - R(1)

which means difference of population of rabbits after 2 months and after 1 month.

$R(1) = 100(\frac{2}{5}.1 )$

$R(1) = 100(\frac{2}{5} )$

R(2) = $100(\frac{2}{5}.2 )$

$R(2) = 100(\frac{4}{5} )$

$\Delta R = 100(\frac{4}{5} )-100(\frac{2}{5} )$

$\Delta R = 100[\frac{4}{5} - \frac{2}{5} ]$

$\Delta R=$ 40

Difference of rabbits between first and second months is 40.

b. R(0) = 100( $\frac{2}{5} .0$ )

R(0) = 0

Initially, there no rabbits in the population.

c. R(10) = $100(\frac{2}{5}.10 )$

R(10) = 400

In 10 months, there will be 400 rabbits.

d. R(t) = 500

$500=100(\frac{2}{5}.t )$

$\frac{500}{100}=\frac{2}{5}.t$

$t = \frac{500.5}{100.2}$

t = 12.5

In 12 and half months, population of rabbits will be 500.

3 0

2 years ago

Simplify the expression c^-6 * d^4/ c^-16 * d^14

Naily [24]

Answer:

$\dfrac{c^{10}}{d^{10}}$

Step-by-step explanation:

Although the way you wrote problem, this is not what it looks like, I think this is what you meant.

$\dfrac{c^{-6} \times d^{4}}{c^{-16} \times d^{14}} =$

$= \dfrac{c^{-6}}{c^{-16}} \times \dfrac{d^{4}}{d^{14}}$

$= c^{-6 - (-16)} \times d^{4 - 14}$

$= c^{-6 + 16} \times d^{-10}$

$= c^{10} \times d^{-10}$

$= \dfrac{c^{10}}{d^{10}}$

7 0

2 years ago

Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0

2 years ago

The point(5/13,Y) IS IN THE 4TH quadrant corresponds to angle on the unit circle

shutvik [7]

4th quadrant is positive x (cos) and negative y (sin)

Use the Pythagorean Theorem to calculate the value of y (sin).
x² + y² = c²
5² + y² = 13²
25 + y² = 169
y² = 144
y = 12

Since the y-value is negative in the 4th quadrant, then y = -12

6 0

3 years ago

Ashley makes house calls. For each she is payed a base amount and makes additional money for each hour she works. The graph belo

Zarrin [17]

Answer:

her pay increases by 30$ each hour she work

Step-by-step explanation:

hope it helps

3 0

1 year ago