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4 years ago

6

During tax-free weekend, Amy bought pencils for $1.25 each. After using a coupon which saved her $3, she spent a total of $2. Ho

w many pencils, x, did Amy purchase?

1 answer:

galina1969 [7]4 years ago

8 0

Answer:

4

Step-by-step explanation:

1.25* 4=5

5-3=2

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What is the total area of the figure below?<br> answer fast pls 20 pts! <3

Reptile [31]

≻───── ⋆✩⋆ ─────≺

Answer:

First, find the area of rectangle.

20 × 30 = 600

Next multiply the square

10 × 10 = 100

Now the curve.

$\frac{10 * 10 = 100}{2}$

100/2 =50

Then Add all of it.

$600 + 100 + 50 = 750$

$Area = 750$

Hope this helps!

-Fishylikeswater

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3 years ago

Which word problem matches the problem? 2 3 x 7 A) Mark has 7 gallons of milk. Ernesto drinks 2 3 of a gallon. How much milk is

kakasveta [241]

Answer:

Out of given word problem , (B) word problem matches the given problem.

B) Mark drinks $\frac{2}{3}$ of a gallon of milk every week. How much does he drink in 7 weeks? written as,

$x=\frac{2}{3}\times 7$

Step-by-step explanation:

Given problem $\frac{2}{3} \times 7$

We are given four options and we have to chose the correct one.

We will convert each given statement into mathematical problem.

Consider,

A)Mark has 7 gallons of milk. Ernesto drinks $\frac{2}{3}$ of a gallon. How much milk is left?

Mathematically represented as,

Let left milk be x then ,

$x=7-\frac{2}{3}$

B)Mark drinks $\frac{2}{3}$ of a gallon of milk every week. How much does he drink in 7 weeks?

Mathematically represented as,

Let milk drink by Mark in 7 weeks be x gallons then ,

$x=\frac{2}{3}\times 7$

C) Mark has $\frac{2}{3}$ of a gallon of milk. Ernesto has 7 gallons of milk. How much do they have together?

Mathematically represented as,

Let they together they drink x gallon milk, then

$x=\frac{2}{3}+ 7$

D) Mark has $\frac{2}{3}$ of a gallon of milk to divide among 7 family members. How much milk does each person get?

Mathematically represented as,

Let each person gets x gallon milk, then

$x=\frac{2}{3} \div 7$

Thus, out of given word problem , (B) word problem matches the given problem

4 0

3 years ago

Read 2 more answers

Write a cosine function that has an amplitude of 2, a midline of 5 and a period of \frac{3\pi}{4} 4 3π .

kakasveta [241]

Given:

Amplitude = 2

Midline = 5

Period of $\dfrac{3\pi}{4}$ .

To find:

The cosine function.

Solution:

The general cosine function is:

$f(x)=A\cos(Bx+C)+D$ ...(i)

Where, |A| is amplitude, $\dfrac{2\pi}{B}$ is period, $-\dfrac{C}{B}$ is phase shift and D is the mid line.

Period of function is $\dfrac{3\pi}{4}$ . So,

$\dfrac{3\pi}{4}=\dfrac{2\pi}{B}$

$B=2\pi\times \dfrac{4}{3\pi}$

$B=\dfrac{8}{3}$

Substituting $A=2,\ B=\dfrac{8}{3},\ C=0,\ D=5$ in (i), we get

$f(x)=2\cos(\dfrac{8}{3}x+0)+5$

$f(x)=2\cos(\dfrac{8}{3}x)+5$

Therefore, the required cosine function is $f(x)=2\cos(\dfrac{8}{3}x)+5$ .

6 0

3 years ago

Find the surface area of the cylinder. Round your answer to the nearest tenth.

xenn [34]

It would be 282.7 or 283 for the total surface area. Since the radius is three and the height is 12.

6 0

3 years ago

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position

stira [4]

Answer:

$v(t) = (2t+1) \mathbb{i} +3t^2 \mathbb{j} +4t^3 \mathbb{k}$

$r(t) = (t^2+t) \mathbb{i} +(t^3+2) \mathbb{j} +(t^4 - 3) \mathbb{k}$

Step-by-step explanation:

The velocity vector is the integral of the acceleration vector i.e.

$v(t) = \int a(t) dt$

$v(t) = \int (2 \mathbb{i}+6t \mathbb{j} 12t^2 \mathbb{k}) dt$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + C_1$

When $t=0$ , $v(0) = \mathbb{i}$ . Inserting these values in $v(t)$ ,

$C_1= \mathbb{i}$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + \mathbb{i}$

$v(t) = (2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}$

The position vector is the integral of the velocity vector i.e.

$r(t) = \int v(t) dt$

$r(t) = \int ((2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}) dt$

$r(t) = (t^2+t) \mathbb{i}+t^3 \mathbb{j} t^4 \mathbb{k} + C_2$

When $t=0$ , $r(t) =2\mathbb{j}-3\mathbb{k}$ . Inserting these values in $r(t)$ ,

$C_2=2\mathbb{j}-3\mathbb{k}$

$r(t)= (t^2+t) \mathbb{i}+t^3 \mathbb{j}+ t^4 \mathbb{k} + 2\mathbb{j}-3\mathbb{k}$

$r(t) = (t^2+t) \mathbb{i}+(t^3+2) \mathbb{j}+ (t^4-3) \mathbb{k}$

5 0

4 years ago