All categories

2 years ago

Find an equation of variation in which y varies directly as x and y=1 when x=0.5. Then find the value of y when x = 14.

Mathematics

1 answer:

FromTheMoon [43]2 years ago

8 0

The value of y is twice that of x, so the equation is y = 2x.

Substituting in x = 14, we get y = 2(14) = 28.

You might be interested in

Find the value of x and y.

Daniel [21]

Good day ~_~ /////////////

6 0

3 years ago

Find the coordinates of the image of point R(3, -5) rotated 180 degrees about the origin.

kolbaska11 [484]

180 degrees means that the new x=-x and the new y=-y so
x=3 and y=-5 is orignal
new is -3 and 5

so new oint is

A(-3,5)

6 0

3 years ago

What is the midpoint of (3,8) and (6,4)

Andrews [41]

(4.5,6) that is what i got.

4 0

3 years ago

Tim typed a 46-word paragraph in 3/4 of a minute. What is his typing speed per minute?

soldier1979 [14.2K]

Answer:i think the answer is something

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Simplify the complex fraction.

lilavasa [31]

Given:

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}$

To find:

The simplified fraction.

Solution:

Step 1: Simplify the numerator

$$\frac{(4 r)^{3}}{15 t^{4}}=\frac{4^3 r^{3}}{15 t^{4}}=\frac{64 r^{3}}{15 t^{4}}$

Step 2: Simplify the denominator

$$\frac{16 r}{(3 t)^{2}}=\frac{16 r}{3^2 t^{2}}= \frac{16 r}{9 t^{2}}$

Step 3: Using step 1 and step 2

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}$

Step 4: Using fraction rule:

$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a \cdot d}{b \cdot c}$

$$\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}}\right)}=\frac{64r^3 \cdot 9t^2}{16 r \cdot 15 t^4}$

Cancel the common factor r and t², we get

$$=\frac{64 r^{2} \cdot 9 }{16 \cdot 15 t^2 }$

Cancel the common factors 16 and 3 on both numerator and denominator.

$$=\frac{4 r^{2} \cdot 3 }{ 5 t^2 }$

$$=\frac{12 r^{2} }{ 5 t^2 }$

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{12 r^{2} }{ 5 t^2 }$

The simplified fraction is $\frac{12 r^{2} }{ 5 t^2 }$ .

5 0

3 years ago