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sergiy2304 [10]

3 years ago

9

The oblique pyramid has a square base.

1 answer:

Brums [2.3K]3 years ago

6 0

Good evening ,

______

Answer:

the volume of the pyramid = 5cm³

___________________

Step-by-step explanation:

the volume of the pyramid = (b×h)/3 = (2²×3,75)/3 = (4×3,75)/3 = 15/3 = 5.

:)

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It would be really helpful if you could solve this for me. Please and thank you:)

katrin2010 [14]

Hey there! :)

Answer:

2nd choice.

Step-by-step explanation:

Given:

630 = 42f.

Divide both sides by 42 to solve for f:

630/42 = 42f/42

f = $15.

This equation is not an inequality, so a line would not be graphed on the number-line. The proper way to show a single solution would be with a point. Therefore:

The second choice is correct. The number line has a single point at f = 15.

3 0

4 years ago

Ms. Winter makes homemade bath soaps and bottles of lotion. In her inventory, she has 48 bath soaps and 64 bottles of lotion. Sh

Levart [38]

She could put together baskets with 8 bath soaps and 8 bottles of lotion in each.

Hope this helps!

5 0

3 years ago

find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2

Alexxandr [17]

The surface is parameterized by

$\vec s(u,v) = x(u, v) \, \vec\imath + y(u, v) \, \vec\jmath + z(u, v)) \, \vec k$

and the normal to the surface is given by the cross product of the partial derivatives of $\vec s$ :

$\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}$

It looks like you're given

$\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}$

Then the normal vector is

$\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k$

Now, the point (2, 2, 0) corresponds to u and v such that

$\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}$

and solving gives $u = v = 1$ , so the normal vector at the point we care about is

$\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k$

Then the equation of the tangent plane is

$\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath + (z-0)\,\vec k\right) = 0$

$-4(x-2) + 2(y-2) - 4z = 0$

$\boxed{2x - y + 2z = 2}$

6 0

2 years ago

HELP PLZ THIS I NEED TO KNOW THIS

Korvikt [17]

ITS BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

7 0

3 years ago

In right △ABC , the right angle is at C, m∠A=30 degrees , and AC=7√5 units.

Archy [21]

Answer:

$P=(7\sqrt{15}+7\sqrt{5})\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the length side AB (hypotenuse)

we know that

$cos(30^o)=\frac{AC}{AB}$

substitute

$cos(30^o)=\frac{7\sqrt{5}}{AB}$

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

so

$\frac{\sqrt{3}}{2}=\frac{7\sqrt{5}}{AB}$

$AB=\frac{14\sqrt{5}}{\sqrt{3}}$

$AB=\frac{14\sqrt{15}}{3}\ units$

step 2

Find the length side BC

$sin(30^o)=\frac{BC}{AB}$

$sin(30^o)=\frac{1}{2}$

$\frac{1}{2}=\frac{BC}{AB}$

$BC=\frac{AB}{2}$

substitute the value of AB

$BC=\frac{\frac{14\sqrt{15}}{3}}{2}$

$BC=\frac{7\sqrt{15}}{3}\ units$

step 3

Find the perimeter

$P=AB+BC+AC$

substitute

$P=\frac{14\sqrt{15}}{3}+\frac{7\sqrt{15}}{3}+7\sqrt{5}$

$P=\frac{21\sqrt{15}}{3}+7\sqrt{5}$

$P=(7\sqrt{15}+7\sqrt{5})\ units$

4 0

3 years ago