What is the height of the wall that is 30 feet long and that required 2310 bricks to build

Please HELPppppppppppppppp

aleksandr82 [10.1K]

Answer:

Point of intersections are (0, -7) and (5, -2).

Step-by-step explanation:

From the graph attached,

A straight line is intersecting the circle at the two points (0, 7) and (5, -2).

Now solve algebraically,

Equation of the line → y = x - 7 -------(1)

Equation of the circle → (x - 5)² + (y + 7)² = 25 -------(2)

By substituting the value of y from equation (1) to equation (2)

(x - 5)² + (x - 7 + 7)² = 25

(x - 5)² + x² = 25

x² - 10x + 25 + x² = 25

2x² - 10x = 0

x² - 5x = 0

x(x - 5) = 0

x = 0, 5

From equation (1),

y = 0 - 7 = -7

y = 5 - 7 = -2

Therefore, point of intersections are (0, -7) and (5, -2).

6 0

3 years ago

Imaginen que un maratonista corre toda una carrera a una velocidad constante. Después de dos horas lleva recorridos 22 km

saw5 [17]

For this case we have:

If after 2 hours the marathon runner has traveled 22 kilometers we have:

2h -----------> 22km

Applying a rule of three, we can know the time it takes to run 42km. So, we have:

2h -----------> 22km

x -------------> 42km

Where "x" is the time it takes to travel 42km. Resolving we have:

$x = \frac {(42 * 2)} {22}$

$x = \frac {84} {22}$

$x = \frac {42} {11}$

$x = 3.82 hours$

Thus, after 3.82 hours the marathon runner will travel 42km.

Answer:

3.82 hours

6 0

2 years ago

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

The area of rectangle is equal to

$A=14*5=70\ in^{2}$

The area of the small semicircle is equal to

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

The area of the larger semicircle is equal to

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

Read 2 more answers

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

PLEASE! ANSWER THIS! HELP! PLEASE!!!!!!!!!!!!!! c. Expand: (tan(theta) - 1)^3

jonny [76]

Answer:

(tan(theta)-1)^3

= (tan(theta)-1)(tan(theta)-1)(tan(theta)-1)

= (tan^2(theta)-2tan(theta)+1)(tan(theta)-1)

= tan^3(theta)-2tan^2(theta)+tan(theta)-tan^2(theta)+2tan(theta)-1

= tan^3(theta)-3tan^2(theta)+3tan(theta)-1

Hope this helps :)

6 0

2 years ago