All categories

3 years ago

What is the unit rate of 13/2 pounds per 3/4 quart?

Mathematics

1 answer:

kodGreya [7K]3 years ago

3 0

Answer:

26/3 or 8.66666 (8.6 with a line over the 6)

Step-by-step explanation:

According to the problem, 13/2 pounds is only 3/4 of a quart. A unit rate, however, has a denominator value of 1. To solve this question, we must change the value of 3/4 to 1.

The first step to this is changing 3/4 to 1/4. This means affecting the value 13/2 pounds. 1/4 is a third of 3/4. We must find a third of 13/2 to properly change this relationship. So, we should do 13/2 ÷ 3. This can also be written as 13/2*3. So, a third of 13/2 is 13/6. Our new relationship is 13/6 pounds = 1/4 quart.

Now, we can multiply 13/6 and 1/4 by 4 to get ? and 4/4, or 1. We can multiply the numerator by 4 to finish this step. 13/6 becomes 13*4/6, which equals 52/6. 52/6 can be divided by 2 to simplify the fraction and becomes 26/3.

So, the <em>unit rate of 13/2 pounds per 3/4 quart is 26/3</em>. 26/3 can also be written as 8.66666, or 8.6 with a line over the 6, in decimal form.

You might be interested in

What is the solution to the equation n + 12 = 17?

yarga [219]

Answer:

n=5

Step-by-step explanation:

12+5=17 is equal to 12+n=17 so n=5

6 0

3 years ago

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x

Andrei [34K]

Answer:

$\large\boxed{A.\ (x+6)^2+(y+10)^2=20}$

Step-by-step explanation:

The standard form of an equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10$

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute (-6, -10) and (-8, -6):

$r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}$

Finally we have

$(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20$

5 0

3 years ago

Read 2 more answers

What is x if 3x-2=x-3

blsea [12.9K]

Answer:

x=-1/2

Step-by-step explanation:

add 2 to both sides

3x=x-1

subtract x from both sides

2x=-1

divide by 2

x=-1/2

7 0

3 years ago

Read 2 more answers

Can u please help me solve? I'm reviewing for finals.

Black_prince [1.1K]

Given:

Consider the given graph as a reference of the solution.

To find:

$-3(u\cdot v)$

Explanation:

By analyzing the graph, we can define the coordinate of vector u and v:

$\begin{align} & \vec{u}=(-8,-9)-(0,0)=(-8,-9) \\ & \vec{v}=(3,7)-(0,0)=(3,7)\end{align}$

Now, let perform the dot product of two vectors,

$\begin{gathered} u\cdot v=(-8,-9)\cdot(3,7) \\ u\cdot v=(-8)(3)+(-9)(7) \\ u\cdot v=-24-63 \\ u\cdot v=-87 \end{gathered}$

Now, perform the required operation,

$\begin{gathered} -3(u\cdot v) \\ =-3(-87) \\ =261 \end{gathered}$

Final answer:

Hence, the required solution is:

$-3(u\cdot v)=261$

8 0

1 year ago

If a varies directly as the cube root of B and if a equals to 3 when B equals to 64 find the formula connecting the variables he

Brrunno [24]

Answer:

a=k1 $\sqrt[3]{B}$

B=125

Step-by-step explanation:

Given :

a=3

B=64

According to question

a ∝ $\sqrt[3]{B}$

therefore

a=k1 $\sqrt[3]{B}$ ........Eq(1)

K1= $\frac{a}{\sqrt[3]{B} }$ ......Eq(2)

Putting the value of a and B we get in Eq(2) we get

$K1=\frac{3}{4}$

Putting the value of k1 in Eq(1)

$a=\frac{3}{4}\sqrt{B}$ .......................Eq(3)

putting the value of a=15/4 IN Eq(3) we get

$\frac{15}{4}\ =\frac{3}{4} \sqrt[3]{B} \\\\\sqrt[3]{B}\ =\ 5\\Cubing\ both\ side\ we\ get\\B=125$

5 0

3 years ago