The area of a rectangle is 15x^2y^5. if the length is 5xy^3, what is the width?

A 10-metre long pipe is cut into 2 pieces. One piece is 5 times the length of the other piece. Find the length of each piece, en

elena55 [62]

9514 1404 393

Answer:

1.67 m, 8.33 m

Step-by-step explanation:

If the ratio of lengths is 1 : 5, then the ratio of the shortest length to the total is ...

1 : (1+5) = 1 : 6

The shortest length is (10 m)×(1/6) = 1.67 m. The longer length is 10 -1.67 = 8.33 m.

The lengths of the pieces are 8.33 m and 1.67 m.

6 0

3 years ago

Lucy’s parents built a swimming pool in the backyard. Use 3.14 for π.

Neporo4naja [7]

Answer:

Distance around the pool = 162.8 feet

Area of the pool = 957 square feet

Step-by-step explanation:

Distance around the swimming pool = Perimeter of the pool

Perimeter of the pool which is a composite figure will be,

= Circumference of the semicircle + Sum of three sides of the pool

= πr + 2×(length of the pool) + width of the pool

= 3.14×(10) + 2×40 + 20

= 62.8 + 80 + 20

= 162.8 ft

Area of the pool = Area of the semicircle + Area of the rectangular pool

= $\frac{1}{2}(\pi)(r)^{2}+(\text{length}\times \text{Width})$

= $\frac{1}{2}(3.14)(10)^2+(40\times 20)$

= 157 + 800

= 957 square feet

8 0

3 years ago

What is the distance between the points (−7, 7) and (−7, 8) ? 1 unit 7 units 8 units 15 units Please Help I Will Mark Brainliest

k0ka [10]

√ 197

≈

14.03566884

So your answer should be 15 units

6 0

3 years ago

Read 2 more answers

Implicit differentiation of 1/x +1/y=5 y(4)= 4/19<br> y'(4)=?

Aneli [31]

$\frac{1}{x} +\frac{1}{y} = 5\\\\x^{-1}+y^{-1}=5\\$

Above, I changed the fraction form of x and y into exponential form so it is easier to see the differentiation. Now, we can differentiate:

$-1x^{-2}+-1y^{-2}\frac{dy}{dx}=5\\\\\frac{-1}{x^2}-\frac{1}{y^2}\frac{dy}{dx}=5\\\\-\frac{1}{y^2}\frac{dy}{dx}=5+\frac{1}{x^2}\\\\\frac{dy}{dx}=-5y^2-\frac{y^2}{x^2}$

Now that we have dy/dx, we can plug in the x, which is 4, and the y, which is 4/19. We know these values of x and y because your question stated y(4) = 4/19.

$\frac{dy}{dx}=-5(\frac{4}{19})^2-\frac{(\frac{4}{19})^2}{(4)^2}\\\\\frac{dy}{dx}=-5(\frac{16}{361})-\frac{(\frac{16}{361})}{16}\\\\\frac{dy}{dx}=\frac{-80}{361}-\frac{1}{361}\\\\\frac{dy}{dx}=\frac{-81}{361}$

5 0

4 years ago

Find the product of the solutions

irina1246 [14]

Answer:

A. -41/15.

Step-by-step explanation:

15x^2 + 23x - 41 = 0

Product of the roots of ax^2 + bx + c = c/a.

Thus the product of the roots of this equation = -41/15.

7 0

3 years ago