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2 years ago

What is the lateral area of the pyramid

Mathematics

1 answer:

Mkey [24]2 years ago

3 0

Answer:

43.75 ft²

Step-by-step explanation:

$s_l$ = (l√(w/2)² + h²) + (w√(l/2)² + h²)

l & w become 3.5, and h becomes 6.

$s_l$ = (3.5√(3.5/2)² + 6²) + (3.5√(3.5/2)² + 6²)

Step 1:Because this is a square pyramid, what you see above essentially becomes what you see below.

$s_l$ = 2(3.5√(3.5/2)² + 6²)

Step 2: Divide 3.5 by 2 to get 1.75.

$s_l$ = 2(3.5√1.75² + 6²)

Step 3: Square both 1.75 and 6 to get 3.0625 and 36 respectively.

$s_l$ = 2(3.5√3.0625 + 36)

Step 4: Add 3.0625 and 36 to get 39.0625.

$s_l$ = 2(3.5√39.0625)

Step 5: The square root of 39.0625 is 6.25.

$s_l$ = 2(3.5 * 6.25)

Step 6: Multiply 3.5 by 6.25 to get 21.875.

$s_l$ = 2(21.875)

Step 7: Multiply 2 by 21.875 to get 43.75.

$s_l$ = 43.75 ft²

The lateral area of this pyramid is 43.75 ft².

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Help with this please

aliina [53]

AB = 6 cm, AC = 12 cm, CD = ?

In triangle ABC, ∠CBA = 90°, therefore in triangle BCD ∠CBD = 90° also.

Since ∠BDC = 55°, ∠CBD = 90°, and there are 180 degrees in a triangle, we know ∠DCB = 180 - 55 - 90 = 35°

In order to find ∠BCA, use the law of sines:

sin(∠BCA)/BA = sin(∠CBA)/CA
sin(∠BCA)/6 cm = sin(90)/12 cm
sin(∠BCA) = 6*(1)/12 = 0.5
∠BCA = arcsin(0.5) = 30° or 150°
We know the sum of all angles in a triangle must be 180°, so we choose the value 30° for ∠BCA

Now add ∠BCA (30°) to ∠DCB = 35° to find ∠DCA.
∠DCA = 30 + 35 = 65°

Since triangle DCA has 180°, we know ∠CAD = 180 - ∠DCA - ∠ADC = 180 - 65 - 55 = 60°

In triangle DCA we now have all three angles and one side, so we can use the law of sines to find the length of DC.

12cm/sin(∠ADC) = DC/sin(∠DCA)
12cm/sin(55°) = DC/sin(60°)
DC = 12cm*sin(60°)/sin(55°)
DC = 12.686 cm

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3 years ago

Can yall help me with this onee??

umka21 [38]

Answer:

Ok so it's been a minute since I've done this however the answer would be A I think.

3x+2y=3
subtract 3x and put it behind 2y
divide by 2 throughout the entire equation
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3 years ago

A person on the ground looks up at an angle of 28° and sees the top of a tree and the top of a building aligned. The tree is 20

lorasvet [3.4K]

Answer:

21.27 meters.

Step-by-step explanation:

Please find the attachment.

Let H and h represent height of building and tree respectively.

We have been given that a person on the ground looks up at an angle of 28° and sees the top of a tree and the top of a building aligned. The tree is 20 m away from the person and the building is 60 m away from the person.

We know that tangent relates opposite side of a right triangle with its adjacent side.

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

$\text{tan}(28^{\circ})=\frac{H}{60}$

$60\cdot \text{tan}(28^{\circ})=H$

$60\cdot 0.531709431661=H$

$H=31.90256589966\approx 31.90$

Similarly, we can find height of the tree.

$\text{tan}(28^{\circ})=\frac{h}{20}$

$20\cdot\text{tan}(28^{\circ})=h$

$20\cdot 0.531709431661=h$

$h=10.63418863322\approx 10.63$

$H-h=31.90-10.63$

$H-h=21.27$

Therefore, the difference in heights between the building and tree is 21.27 meters.

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For the given function, find the vertical and horizontal asymptotes f(x)= x2+3/ x2-9

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