Ask question

Ask question

All categories

2 years ago

5

Zoe compares the price of a sweatshirt at two stores. At Store A, the sweatshirt costs $25. A friend told her that the sweatshir

t costs 30% more at Store B.
Zoe wrote the expression 25+0.3(25) to find the price of the sweatshirt at Store B.

Which equivalent expression could she also use to find the price of the sweatshirt at Store B?

A) (1+0.3)25
B) (1−0.3)25
C) 1+0.3(25)
D) 25 + 0.3

1 answer:

Rudiy272 years ago

8 0

Answer:

<u>A) (1+0.3)25</u>

<u />

Explanation:

Zoe's expression <u>25+0.3(25)</u> equals to 32.5. So the only expression that equals to 32.5 is <u>A) (1+0.3)25</u>.

You might be interested in

The height y (in feet) of a toss in bocce ball can be modeled by y=-x^2+4x, where x is the horizontal distance (in feet)

Ber [7]

4 feet, the X-intercept tells us the value of X, aka how far the ball landed from 0

6 0

3 years ago

How can you solve an equation without using the distributive property

belka [17]

You could use a calculator.. or if it is already in distributive property then just add it up

6 0

2 years ago

A dilation with a scale factor of 3.5 and centered at the origin is applied to PQ¯¯¯¯¯ with endpoints P(2, 5) and Q(2, 1).

pshichka [43]

The answer for p is (7,17.5)
Q is (7,3.5) I have the same test

5 0

2 years ago

Read 2 more answers

"Dan" uses 2/3 cup of brown sugar and 1/4 cup of white sugar to make oatmeal bars. How many cups of sugar does he use in all?

hammer [34]

Answer:

the number of cups of sugar he use in all is $\frac{11}{12}$

Step-by-step explanation:

The number of cups does he use in all as follows:

= Brown sugar + white sugar

$= \frac{2}{3} + \frac{1}{4}\\\\= \frac{8 + 3}{12} \\\\= \frac{11}{12}$

Hence, the number of cups of sugar he use in all is $\frac{11}{12}$

8 0

2 years ago

1. (1 pt each) The equation of motion of a particle is s = t 3 − 12t 2 + 36t, t ≥ 0, where s is measured in meters and t is in s

Inga [223]

QUESTION 1)

The particle's equation of motion is

$s=t^3-12t^2+36t,t\ge0$

where s is measured in meters and t is in seconds.

The velocity at time, $t$ is given by the first derivative of the equation of motion of the particle.

$s'(t)=(3t^2-24t+36)ms^{-1}$

Question 1b).

To find the velocity after 3 seconds, we put $t=3$ into the velocity function.

$\Rightarrow s'(3)=(3(3)^2-24(3)+36)ms^{-1}$

$\Rightarrow s'(3)=(27-72+36)ms^{-1}$

$\Rightarrow s'(3)=-9ms^{-1}$

QUESTION 1C

To find the time that the particle is at rest, we equate the velocity (the first derivative formula) to zero and solve for t.

$\Rightarrow 3t^2-24t+36=0$

Divide through by 3 to get;

$\Rightarrow t^2-8t+12=0$

Factor:

$\Rightarrow (t-2)(t-6)=0$

$(t-2)=0,(t-6)=0$

$t=2s,t=6s$

The particle is at rest when $t=2s$ and $t=6s$ .

QUESTION 1d.

The particle is moving in a positive direction when the velocity is greater than zero.

$\Rightarrow 3t^2-24t+36\:>\:0$

$\Rightarrow t^2-8t+12\:>\:0$

$\Rightarrow (t-2)(t-6)\:>\:0$

$\Rightarrow t\:\:6$

But $t\ge0$ .

This implies that, the particle is moving in a positive direction on the interval,

$0\le t\:\:6$

QUESTION 1e.

To find the total distance traveled after 8s, we substitute $t=8$ into the equation of motion of the particle.

$s(8)=8^3-12(8)^2+36(8)$

$s(8)=512-768+288$

$s(8)=32m$

The particle covered 32m in the first 8 seconds.

QUESTION 1f

i) The acceleration at time $t$ can be obtained by differentiating the velocity equation.

$\Rightarrow s"(t)=6t-24$

ii) To find the acceleration after 3 seconds, we substitute $t=3$ into the equation of acceleration.

$\Rightarrow s"(3)=6(3)-24$

$\Rightarrow s"(3)=18-24$

$\Rightarrow s"(3)=-6ms^{-2}$

After 3 seconds, the particle is decelerating at 6 meters per seconds square.

QUESTION 2a

Given:

$p(x)=x^n-x$

$p'(x)=nx^{n-1}-1$

When $n=2$ , then

$p'(x)=2x^{2-1}-1$

$p'(x)=2x-1$

This implies that, p(x) is decreasing when

$p'(x)\:$

$2x-1\:$

Therefore the function is decreasing on;

$x\:$

QUESTION 2b

When $n=\frac{1}{2}$

$p'(x)=\frac{1}{2\sqrt{x}}-1$

To find the interval over which the function is decreasing, we solve the inequality;

$\frac{1}{2\sqrt{x}}-1\:$

Therefore the function is decreasing on the interval;

$\Rightarrow x\:>\:\frac{1}{4}$

QUESTION 3a

The given parabola has equation

$y=x^2+x...(1)$

Let the two tangents from the external point; $(2,-3)$ have equation;

$y+3=m(x-2)..(2)$

Put equation (1) into equation (2)

This implies that;

$x^2+x+3=m(x-2)$

Rewrite to obtain a quadratic equation in $x$ .

$x^2+(1-m)x+3+2m=0$

Since this is the point of intersection of a tangent and a parabola, the discriminant of this quadratic equation must be zero.

$\Rightarrow (1-m)^2-4(3-2m)=0$

$\Rightarrow m^2-10m-11=0$

$\Rightarrow m=11\:or\:m=-1$

We substitute the values of m into equation (2) to obtain the equations of the two tangents to be;

$y=11x-25$ and $y=-x-1$

QUESTION 3b

Let the equation of the tangents from the external point (2,7) be

$y-7=m(x-2)...(1)$

The given parabola has equation

$y=x^2+x...(2)$

The discriminant of the intersection of these two equations yields;

$m^2-10m+29=0$

This quadratic equation has no real roots.

Hence there are no lines through the point (2,7) that are tangents to parabola.

We can see from the graph that this point is lying inside the parabola.

QUESTION 4

The given cubic function is

$y=ax^3+bx^2+cx+d$

$\frac{dy}{dx}=3ax^2+2bx+c$

The horizontal tangents occurs when $\frac{dy}{dx}=0$ .

$\Rightarrow 3ax^2+2bx+c=0$

This occurs at (2,0).

$\Rightarrow 12a+4b+c=0...(1)$

and (-2,6) .

$\Rightarrow 12a-4b+c=0...(2)$

These points also lie on the curve so they must satisfy the equation of the curve;

Substituting (2,0) into the original equation gives;

$8a+4b+2c+d=0...(3)$

Substituting (-2,6) into the original equation gives;

$-8a+4b-2c+d=0...(4)$

Solving the four equations simultaneously gives;

$a=\frac{3}{16},b=0,c=-\frac{9}{4},c=3$

Hence the required cubic function;

$y=\frac{3}{16}x^3-\frac{9}{4}x+3$

QUESTION 5a

Let

$y=fgh$

where $f,g,$ and $h$ are differentiable.

Using the product rule;

$y'=f'(gh)+f(gh)'$

Use the product rule again;

$y'=f'(gh)+f(g'h+gh')$

$y'=f'(gh)+fg'h+fgh'$ as required.

3 0

2 years ago