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At a fundraiser, students sold chocolate bars with almonds and chocolate bars with walnuts. The number of chocolate bars with al

BabaBlast [244]

Answer:

Total number chocolate bars with almonds sold were approximately 66.34 and total number chocolate bars with walnuts sold were approximately 34.67.

Step-by-step explanation:

Let number chocolate bars with almonds be x

Also let number chocolate bars with walnuts be y

Given:

Number of chocolate bars with almonds that were sold one weekend was 3 less than 2 times the number of chocolate bars with walnuts.

Framing the above sentence in equation form we get;

$x=2y -3$

Also given:

The number of chocolate bars with walnuts plus 4 times the number of chocolate bars with almonds was 300.

Framing the above sentence in equation form we get;

$4x+y=300$

Now Substituting the value of x in above equation we get;

$4x+y=300\\4(2y-3)+y=300\\8y-12+y=300\\9y=300+12\\9y=312\\\\y=\frac{312}{9} \approx 34.67$

Now we will substitute the value of y in equation $x=2y -3$ we get;

$x=2y-3 = 2\times34.67 -3 = 69.34-3\approx66.34$

Hence, Total number chocolate bars with almonds sold were approximately 66.34 and total number chocolate bars with walnuts sold were approximately 34.67.

6 0

3 years ago

Angela wrote the number 186,425 on the board in which number is the value of the digit six exactly 10 times the value of the dig

Butoxors [25]

So what’s the question? What are we trying to find?

4 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

Which point lies on the line with point-slope equation y - 3 = 4(x + 7)?

Oksanka [162]

Answer:

D.

(-7, 3)

Step-by-step explanation:

Given equation

y - 3 = 4(x + 7)

points are in form of (x,y) that is first value represent value of coordinate and

second value y represents value of y coordinates.

to find which point lies on given line equation we need to substitute value of x and see if value of y obtained is same as given in the option or not.

______________________________________

A.

(7, 3)

substituting x with 7 in the equation

y - 3 = 4(7 + 7)

=> y - 3 = 4(14)

=> y - 3 = 56

=> y = 56+3 = 59

Thus, we see value of y when x is 7 is 59

which is not equal to 3 as given in option A hence option A is not correct choice

B.

(7, -3)

As seen above for x = 7 , y is 59 hence option B is also incorrect

C.

(-7, -3)

substituting x with -7 in the equation

y - 3 = 4(-7 + 7)

=> y - 3 = 4(0)

=> y - 3 = 0

=> y = 0+3 = 3

Thus, we see value of y when x is -7 is 3

which is not equal to -3 as given in option C hence option C is not correct choice

D.

(-7, 3)

As calculated above value of y when x is -7 is 3

hence

option D is correct choice.

8 0

3 years ago

What is an equation for the linear function whose graph contains the points (2, 5) and (3, 0) ?

S_A_V [24]

Y = -5x + 15

You can use point slope formula to get the answer

5 0

3 years ago

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