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2 years ago

1: 18 is __ % of 30

Mathematics

1 answer:

Zigmanuir [339]2 years ago

4 0

Answer:

1. 60%

2. 70

3. 8%

Step-by-step explanation:

I attached pictures to show my work (sorry if they’re a bit messy)!

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ASAP: I need as soon as possible with this question I have taken a picture of. If you could tell me how to do it and the answer,

VladimirAG [237]

Answer:

The answer to your question is h = 125.85 ft = 126 ft

Step-by-step explanation:

Process

1.- Determine two equations to solve the problem

$tan 14 = \frac{h}{x}$ (1)

$tan 47 = \frac{h}{x - 386}$ (2)

from (1) h = xtan14

substitute in (2) tan 47 = $\frac{xtan14}{x - 386}$

solve for x tan47(x - 386) = xtan14

1.07x - 413.9 = 0.25x

1.07x - 0.25x = 413.9

0.82x = 413.9

x = 413.9/0.82

x = 504.76 ft

2.- Calculate h

h = 504.76 tan 14

h = 125.85 ft = 126 ft

4 0

4 years ago

What is the surface area of the right trapezoidal prism? To receive credit, you must show the work used to arrive at a final ans

julia-pushkina [17]

Answer:

210 cm²

Step-by-step explanation:

The net of the right trapezoidal prism consists of 2 trapezoid base and four rectangles.

Surface area of the trapezoidal prism = 2(area of trapezoid base) + area of the 4 rectangles

✔️Area of the 2 trapezoid bases:

Area = 2(½(a + b)×h)

Where,

a = 7 cm

b = 11 cm

h = 3 cm

Plug in the values

Area = 2(½(7 + 11)×3)

= (18 × 3)

Area of the 2 trapezoid bases = 54 cm²

✔️Area of Rectangle 1:

Length = 6 cm

Width = 3 cm

Area = 6 × 3 = 18 cm²

✔️Area of Rectangle 2:

Length = 7 cm

Width = 6 cm

Area = 7 × 6 = 42 cm²

✔️Area of Rectangle 3:

Length = 6 cm

Width = 5 cm

Area = 6 × 5 = 30 cm²

✔️Area of Rectangle 4:

Length = 11 cm

Width = 6 cm

Area = 11 × 6 = 66 cm²

✅Surface area of the trapezoidal prism = 54 + 18 + 42 + 30 + 66 = 210 cm²

7 0

3 years ago

.109 is what in fraction form?

dezoksy [38]

Answer: 109/1000

.109 can not be reduced, so we must put it in a fraction with 1,000.

So, the fraction would be 109/1000, because we need a denominator as 1,000.

The answer = 109/1000

6 0

3 years ago

Consider the equation below.

Korvikt [17]

Answer:

Equation in square form:

$y=3(x+5)^2-4$

Extreme value:

$(h,k)=(-5,-4)$

Step-by-step explanation:

We are given

$y=3x^2+30x+71$

we can complete square

$y=3(x^2+10x)+71$

we can use formula

$a^2+2ab+b^2=(a+b)^2$

$y=3(x^2+2\times x\times 5)+71$

now, we can add and subtract 5^2

$y=3(x^2+2\times x\times 5+5^2-5^2)+71$

$y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71$

$y=3(x+5)^2-75+71$

So, we get equation as

$y=3(x+5)^2-4$

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

$y=a(x-h)^2+k$

where

vertex=(h,k)

now, we can compare and find h and k

$y=3(x+5)^2-4$

we get

h=-5

k=-4

so, extreme value of this equation is

$(h,k)=(-5,-4)$

6 0

4 years ago

Read 2 more answers

Revenue

denis23 [38]

Answer:

Step-by-step explanation:

Given that a small business assumes that the demand function for one of its new products can be modeled by

$p=ce^{kx}$

Substitute the given values for p and x to get two equations in c and k

$40 = ce^{1200k} \\45 = ce^{1000k}$

Dividing on by other we get

$\frac{45}{40} =e^{-200k} \\-200 k = ln (45/40) = 0.117583\\k = -0.000589$

Substitute value of k in any one equation

$45 = ce^{-0.589} \\c=45.02651$

b) Revenue of the product is demand and price

i.e. R(x) = p*x = $45.02651xe^{-0.000589x}$

Use Calculus derivative test to find max Revenue

R'(x) = $45.02651 e^{-0.000589x}-45.02651*0.000589 x e^{-0.000589x}\\$

EquateI derivative to 0

1-0.000589x =0

x = 1698.037

When x = 1698 and p = 16.56469

7 0

3 years ago