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nordsb [41]
3 years ago
7

Find the surface area of the right square pyramid. Round your answer to the nearest hundredth.

Mathematics
1 answer:
jekas [21]3 years ago
7 0

Answer:

A. 176

Step-by-step explanation: First, to find the surface area to each triangle, you multiply the base times the height by 1/2. It is 28 for each triangle. Multiply that by 4, and you get 112. Then, the surface area of the square on bottom is 64. When added together, you get 176. No rounding needed. Hope it helped and is correct!

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Three children together have 50 toys,Harry has 10 less toys than Mark. Sue has twice as many toys as Harry. How many toys does e
Alexeev081 [22]

Answer:

Harry = 10 toys

Mark = 20 toys

Sue = 20 toys

Step-by-step explanation:

H + M + S = 50

H = M - 10

S = 2H

Harry = 10 toys

Mark = 20 toys

Sue = 20 toys

5 0
3 years ago
Triangle ABC with A at 0 comma 0, B at 2 comma 4 and C at 0 comma 2, the measure of angle C is 127 degrees, and the measure of a
Nikolay [14]

Answer:

The Answer is B.

Step-by-step explanation:

I got it right on my test

For further evidence here is a picture of the answer

6 0
2 years ago
Reporting Category: Geometry Standard: 10 G.3 Recognize and solve problems involving angles formed by transversals of coplanar l
Alexxx [7]
The measure in degrees would be 15<span />
4 0
3 years ago
Can yall help me out real quick. unit 6 similar triangles homework 3 proving triangles similar. 65 points for this question. Rea
Vesna [10]

Answer:

1, 2 and 7

Step-by-step explanation:

they look the same, have a good day!!

5 0
3 years ago
Help. <br>Please its urgent show workings.<br>​
laiz [17]

Answer:

see explanation

Step-by-step explanation:

There are 2 possible approaches to differentiating these.

Expand the factors and differentiate term by term, or

Use the product rule for differentiation.

I feel they are looking for use of product rule.

Given

y = f(x). g(x) , then

\frac{dy}{dx} = f(x).g'(x) + g(x).f'(x) ← product rule

(a)

y = (2x - 1)(x + 4)²

f(x) = 2x - 1 ⇒ f'(x) = 2

g(x) = (x + 4)²

g'(x) = 2(x + 4) × \frac{d}{dx} (x + 4) ← chain rule

       = 2(x + 4) × 1

        = 2(x + 4)

Then

\frac{dy}{dx} = (2x - 1). 2(x + 4) + (x + 4)². 2

    = 2(2x - 1)(x + 4) + 2(x + 4)² ← factor out 2(x + 4) from each term

    = 2(x + 4) (2x - 1 + x + 4)

    = 2(x + 4)(3x + 3) ← factor out 3

    = 6(x + 4)(x + 1)

--------------------------------------------------------------------------

(b)

y =  x(x² - 1)³

f(x) = x ⇒ f'(x) = 1

g(x) = (x² - 1)³

g'(x) = 3(x² - 1)² × \frac{d}{dx} (x² - 1) ← chain rule

        = 3(x² - 1)² × 2x

        = 6x(x² - 1)²

Then

\frac{dy}{dx} = x. 6x(x² - 1)² + (x² - 1)³. 1

    = 6x²(x² - 1)² + (x² - 1)³ ← factor out (x² - 1)²

    = (x² - 1)² (6x² + x² - 1)

     = (x² - 1)²(7x² - 1)

----------------------------------------------------------------------

(c)

y = (x² - 1)(x³ + 1)

f(x) = x² - 1 ⇒ f'(x) = 2x

g(x) = (x³ + 1) ⇒ g'(x) = 3x²

Then

\frac{dy}{dx} = (x² - 1). 3x² + (x³ + 1), 2x

   = 3x²(x² - 1) + 2x(x³ + 1) ← factor out x

   = x[3x(x² - 1) + 2(x³ + 1) ]

   = x(3x³ - 3x + 2x³ + 2)

   = x(5x³ - 3x + 2) ← distribute

    = 5x^{4} - 3x² + 2x

--------------------------------------------------------------------

(d)

y = 3x³(x² + 4)²

f(x) = 3x³ ⇒ f'(x) = 9x²

g(x) = (x² + 4)²

g'(x) = 2(x² + 4) × \frac{d}{dx}(x² + 4) ← chain rule

       = 2(x² + 4) × 2x

       = 4x(x² + 4)

Then

\frac{dy}{dx} = 3x³. 4x(x² + 4) + (x² + 4)². 9x²

    = 12x^{4}(x² + 4) + 9x²(x² + 4)² ← factor out 3x²(x² + 4)

    = 3x²(x² + 4) [ 4x² + 3(x² + 4) ]

    = 3x²(x² + 4)(4x² + 3x² + 12)

    = 3x²(x² + 4)(7x² + 12)

5 0
2 years ago
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