All categories

3 years ago

A 9 foot ladder is placed 4 feet from the base of a building.about how high does the ladder reach

Mathematics

1 answer:

NeTakaya3 years ago

7 0

Answer:

8.06 feet

Step-by-step explanation:

use the Pythagorean theorem

9^2 = 4^2 + x^2

81 =16 +x^2

x^2 = 65

x = 8.06 feet

You might be interested in

Simplify. 5 2+20 5 30 45 72 (The 5 2 is an exponent.)

nekit [7.7K]

Answer:C.45

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

(1,-4), (2,1), (3,4), (3,3), (4,2) is this a function?

djverab [1.8K]

no i think because theres a negative

3 0

3 years ago

Jesse sold cupcakes and cookies yesterday. Each cupcake sold for $1.50 and each cookie sold for $0.25. At the end of the day, Je

ale4655 [162]

Answer: 19 cupcakes and 37 cookies were sold.

Step-by-step explanation:

Let x represent the number of cupcakes that Jesse sold yesterday.

Let y represent the number of cookies that Jesse sold yesterday.

If he sold 56 cupcakes and cookies combined, it means that

x + y = 56

Each cupcake sold for $1.50 and each cookie sold for $0.25. At the end of the day, Jesse had sold $37.75 worth of cookies and cupcakes. This means that

1.5x + 0.25y = 37.75 - - - - - - - - - - -1

Substituting x = 56 - y into equation 1, it becomes

1.5(56 - y) + 0.25y = 37.75

84 - 1.5y + 0.25y = 37.75

- 1.5y + 0.25y = 37.75 - 84

- 1.25y = - 46.25

y = - 46.25/ - 1.25

y = 37

x = 56 - y = 56 - 37

x = 19

7 0

3 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

2 years ago

What is an equation of the line that passes through the point (-2,-4) and is parallel to the line 5x-2y=6?

Goshia [24]

Answer:

well the answer is still 6 I could help more if you want but I have to work on appreciation

Step-by-step explanation:

on 2 level

8 0

3 years ago