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3 years ago

A) Which functions have graphs with y-intercepts greater than 4?

Mathematics

1 answer:

arlik [135]3 years ago

5 0

Answer:

below

Step-by-step explanation:

fubction 1: ←Function 1 has the

y-intercept- (2)  greatest slope

Slope- (3) 

function 2: ← function with the y-intercept

y-intercept- (5) greater than 4

Slope- (2)

Function 3: ← Function 3 has the y-intercept

Y-intercept- (-1) closest to zero

Slope- (1)

function 4:

y-intercept- (-4)

Slope- (-5)

Hoped that helped:P

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478 × 965 in distributive property

kakasveta [241]

478(900) + 478(95)

Hope this helps!

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4 years ago

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Two urns contain white balls and yellow balls. The first urn contains 2 white balls and 7 yellow balls and the second urn contai

Romashka [77]

-- The first urn has 9 balls in it all together, and 2 of them are white.
If you don't peek, then the prob of pulling out a white ball is 2/9 .

-- The second urn has 13 balls in it all together, and 3 of them are white.
If you don't peek, then the prob of pulling out a white ball is 3/13 .

-- The probability of being successful BOTH times is

(2/9) x (3/13) = ( 6/117 ) = about 0.0513 or 5.13% (rounded)

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4 years ago

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Equation editor does not include the grouping symbols “<“ and “>” that are necessary for writing a vector in component for

Scilla [17]

Answer:

{12,2}

Step-by-step explanation:

From the given graph it is clear that the initial point of the vector is (-5,0) and the terminal point (7,2).

If initial point of a vector is $(x_1,y_1)$ and terminal point is $(x_2,y_2)$ , then

$\overrightarrow{v}=(x_2-x_1)i+(y_2-y_1)j$

Using this formula, we get

$\overrightarrow{v}=(7-(-5))i+(2-0)j$

$\overrightarrow{v}=12i+2j$

$\overrightarrow{v}=$

Using braces, we get

$\overrightarrow{v}=\{12,2\}$

Therefore, the required vector is {12,2}.

4 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

The sector COB is cut from the circle with center O. The ratio of the area of the sector removed from the whole circle to the ar

mafiozo [28]

Answer:

$Ratio = \frac{R^2 - r^2 }{ r^2}$

Step-by-step explanation:

Given

See attachment for circles

Required

Ratio of the outer sector to inner sector

The area of a sector is:

$Area = \frac{\theta}{360}\pi r^2$

For the inner circle

$r \to radius$

The sector of the inner circle has the following area

$A_1 = \frac{\theta}{360}\pi r^2$

For the whole circle

$R \to Radius$

The sector of the outer sector has the following area

$A_2 = \frac{\theta}{360}\pi (R^2 - r^2)$

So, the ratio of the outer sector to the inner sector is:

$Ratio = A_2 : A_1$

$Ratio = \frac{\theta}{360}\pi (R^2 - r^2) : \frac{\theta}{360}\pi r^2$

Cancel out common factor

$Ratio = R^2 - r^2 : r^2$

Express as fraction

$Ratio = \frac{R^2 - r^2 }{ r^2}$

6 0

3 years ago