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Vanyuwa [196]
3 years ago
12

At one store,5 pairs of jeans and 2 sweatshirts cost $208,while 3 pairs of jeans and 4 sweatshirts cost $178.Find the cost of on

e sweatshirt.
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

$19

Step-by-step explanation:

the following equations can be determined from the question

5j + 2s = 208 equation 1

3j + 4s = 178  equation 2

where j = jeans

s = sweatshirts

simultaneous equation would be used to determine the values

Multiply equation 1 by 2

10j + 4s = 416 equation 3

subtract equation 2 from 3

7j = 238

j= $34

substitute for j in equation 1

5(34) + 2s = 208

2s = 38

s = $19

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Answer:

Area of the Trapezoid is=  13.5cm^2

Step-by-step explanation:

The given figure is in the shape of a trapezoid:

Area of a trapezoid= (1/2)*(Sum of parallel opposite sides)* (Distance between the parallel sides)

The parallel sides are: 6 cm and 3 cm

Distance between parallel sides= 3 cm

Area:

                    =\frac{1}{2} *(6+3)*3\\\\=\frac{1}{2}*9*3\\\\=\frac{27}{2} \\\\=13.5 cm^2

Area of the Trapezoid is=  13.5cm^2

5 0
3 years ago
First you to find the worksheet and download it<br> plase I need help
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)

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judy got 117 problems right on a math test in her night school class. She got a score of 78% right. How many problems were on th
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Answer:

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Step-by-step explanation: I hope it helps :)

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