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3 years ago

The function varies directly with x and when x = 30.

Mathematics

2 answers:

Hoochie [10]3 years ago

7 0

Answer:

$f(6)=18$

In words, the image assigned to $x=6$ is 18.

Step-by-step explanation:

Givens

The function varies directly.
When $x=30$ , $f(x)=90$ .

When a function varies directly, it means that function is linear, and it has a direct proportion between variables, so can be modeled by $f(x)=mx$ , where $m$ is the constant ratio of change.

So, in this case, the ratio of change is

$m=\frac{\Delta y}{\Delta x}=\frac{90}{30}=3$

That means the linear function that represents this situation is

$f(x)=3x$

So, if $x=6$ , then

$f(x)=3x\\f(6)=3(6)=18$

Therefore, $f(6)=18$

kap26 [50]3 years ago

3 0

The function f(x) varies directly with x and f(x) = 90 when x = 30.

f(x)=3x

What is f(x) when x = 6?

f(6)= 3*6 =18

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PtichkaEL [24]

Answer:

$IJ=14.8$

Step-by-step explanation:

In a triangle the sum of all angle measures is always 180 degrees

$A+B+C=180$

$2(67)+C=180$

$C=46$ degrees

By splitting the isosceles triangle down the middle we can calculate the base

So ∠H = 23°, ∠K = 90°, ∠J = 67°, HJ = 19

With this we can solve for half of the base. the trig function will need to use is sin, since the side opposite of ∠H will be part of the base. $sin=\frac{opposite}{hypotenuse}$

now we need to rearrange the formula for what we have.

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3 years ago

Timed please i will pick brainliest

iragen [17]

let

x---------> Aviva’s age

y--------->Kanti’s age

z--------> Lakshmi’s age

we know that

y=x-3-----> equation 1

z=2*y-----> equation 2

substitute equation 1 in equation 2

z=2*(x-3)

Aviva’s age------> x

Kanti’s age-------> x-3

Lakshmi’s age----> 2*(x-3)

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the answer is the option

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m_a_m_a [10]

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Veseljchak [2.6K]

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So, the correspondant system

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will look like this:

$\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]$

This turn into the following system of equations:

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The last equation is solvable for $x_4$ : we easily have

$x_4=\dfrac{b_4}{a_{4,4}}$

Once the value for $x_4$ is known, we can solve the third equation for $x_3$ :

$x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}$

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