All categories

2 years ago

Question 13 Help please

Mathematics

1 answer:

Nitella [24]2 years ago

4 0

9514 1404 393

Answer:

f(x) = 11x -15

Step-by-step explanation:

Often, we let x represent the input to a function. When that is multiplied by 11, the result is represented by 11x. If 15 is subtracted from that, then the expression becomes ...

11x -15

If this value is the output of function f, we write this as ...

f(x) = 11x -15

You might be interested in

Multiply (3x + 2) (4x-5)

Rama09 [41]

Answer:

12x^2-7x-10

Step-by-step explanation:

I did binomial into binomial method

3x(4x-5)+2(4x-5)

12x^2-15x+8x-10

12x^2+(-15+8)x-10

12x^2-7x-10

6 0

2 years ago

Read 2 more answers

If jill and katie share 42 chocolates in the ratio of 3:4 then how may chocolates does jill get?

nasty-shy [4]

Answer:

jil got 18 chocolates

Step-by-step explanation:

let the ratio constants be x,

so, jil has 3x chocolates

katie has 4x chocolates

and we know total chocolates = 42

then,

=》3x + 4x = 42

=》7x = 42

=》x = 42 ÷ 7

=》x = 6

so, jil got 3x = (3 × 6) = 18 chocolates

katie got 4x = (4 × 6) = 24 chocolates

5 0

2 years ago

I will mark brainliest please help me on this question!

Tanzania [10]

It’s option a I have done the assignment myself

5 0

3 years ago

What is g(–3) in the function g(x) = 4x2 + 3?

EastWind [94]

Answer:

x is equal to -3. So it would be g(-3)=4(-3)+3 and the solve that.

Step-by-step explanation:

Hope this is helpful

4 0

3 years ago

A car is moving with a velocity (3.0 m/s) x̂ + (1.0 m/s) ŷ and 3.0 seconds later its velocity is (6.0 m/s) x̂ - (3.0 m/s) ŷ. Wha

Mars2501 [29]

Here we know that the initial velocity of the car is given by:

$V_i=(3.0\hat x+1.0\hat y)m/s$

And the final velocity of the car is given by:

$V_f=(6.0\hat x-3.0\hat y)m/s$

It took 3 seconds to attain the final velocity, so we have $t=3 s$

Therefore, the acceleration can be obtained by:

$V_f-V_i=a\times t$

$a=\frac{V_f-V_i}{t}$

Plugging the values of the initial, final velocity and the time, we get:

$a= \frac{(6\hat x-3\hat y)-(3 \hat x+1 \hat y)}{3} =\frac{6\hat x-3 \hat y -3 \hat x-1 \hat y}{3} =\frac{3 \hat x-4 \hat y}{3}$

So the acceleration of the car is given by:

$a=\frac{3 \hat x}{3} -\frac{4 \hat y}{3} =1 \hat x- \frac{4}{3} \hat y$

Now we need to find the direction of the average acceleration of the car:

$\theta=tan^{-1} \frac{y}{x}$

Here, x and y are the coefficients of the 'x' and 'y' components of the vector:

$\theta=tan^{-1}(\frac{-\frac{4}{3} }{1} )=tan^{-1}(\frac{-4}{3}) =-53.13^\circ$

Therefore, the direction of the average acceleration of the car is $-53.13^\circ$ .

8 0

3 years ago