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mamaluj [8]
2 years ago
9

What is the surface area of the figure?

Mathematics
1 answer:
Sphinxa [80]2 years ago
4 0

Answer:

The answer is 24cm²

Step-by-step explanation:

The triangle area is given by A*½ and we get 4*2*½=6

Rectangle one is given as L*W and we solve 3*4= 12

Rectangle two is solved as L*W as well so we solved 1.5*4= 6

and we get 24cm²

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Marcus' employee benefits include family health care coverage. He contributes 25% of the cost. Marcus gets paid weekly and $102
ArbitrLikvidat [17]

Monthly amount, M = $102×4 = $408.

Also, yearly amount, Y = M×12 = $408×12 = $4896

It is given that 25% of the cost is contributed by Marcus.

Amount Marcus paid

=\dfrac{\$4896 \times 25 }{100}\\\\= \$1224

Therefore, his employer contribution annually for coverage is $( 4896 - 1224)=$3672.

Hence, this is the required solution.

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3 years ago
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Need the answer for this angle measures<br> A? <br> B (4x-16)<br> C (5x+7)
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Solve the equation below <br> 9 = x/18 <br> x=<br> What is X?
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7 0
2 years ago
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Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

7 0
2 years ago
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