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Andrews [41]
3 years ago
9

If Anne's age is represented by the variable (a) then.... Epress Jenny's age in terms of (a) if Jenny is 3 years less than 1/2 o

f Anne's age.
Express Fraz's age in terms of (a) if Fraz is twice the sum of Anne's age and 7
Mathematics
1 answer:
Alchen [17]3 years ago
5 0
Jenny is 3 less than half of a, or 1/2a - 3.  Fraz is twice a + 7, or 2a+7.
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A bakery uses ounces of icing for every of a cake. What is the unit rate in ounces of icing per cake?
mr_godi [17]

Answer:

<u>The unit rate in ounces of icing per cake is 40 4/5 or 204/5</u>

Complete statement and question:

A bakery uses 10 1/5 ounces of icing for every 1/4 of a cake. What is the unit rate in ounces of icing per cake

Source: A previous question found at brainly

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Size of icing for every 1/4 of a cake = 10 1/5 ounces or 51/5 ounces

2. What is the unit rate in ounces of icing per cake?

For answering the question, we will use the Rule of Three Simple, this way:

Unit rate in ounces of icing per cake * 1/4 of a cake = Total cake * Unit rate in ounces of icing per 1/4 of a cake

Replacing with the real values, we have:

Unit rate in ounces of icing per cake * 1/4 = 1 * 51/5

Unit rate in ounces of icing per cake = 51/5 / 1/4

Unit rate in ounces of icing per cake = 51/5 * 4/1

Unit rate in ounces of icing per cake = 204/5 ounces or 40 4/5 ounces

<u>The unit rate in ounces of icing per cake is 40 4/5 or 204/5</u>

7 0
3 years ago
Which of these numbers has the greatest absolute value? 1,5,-3,-9
Neko [114]
The -9 because absolute value is the number itself without positive or negative. So it’s the 9. Hope this helps!
4 0
2 years ago
Read 2 more answers
Please help me. After charging for 1/3 of an hour, a phone is at 2/5 of its full power. How long will it take the phone to charg
yanalaym [24]

Answer:

D

Step-by-step explanation:

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)

7 0
2 years ago
How to find the point slope form with the slope of 3 and the point of -2/6
Levart [38]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:y-6 = 3(x + 2)

____________________________________

\large \tt Solution  \: :

Equation of line in point - slope form :

\qquad \tt \rightarrow \:y-y_1 = m(x - x_1)

  • \tt  \:m = slope=3

  • \tt y_1 = y \:  \: coordinate \:  \: of \:  \: point = 6

  • \tt  \:x_1 = x \:  \: coordinate \:  \: of \:  \: point =  - 2

Here :

\qquad \tt \rightarrow \:y - 6 = 3(x -  ( - 2))

\qquad \tt \rightarrow \:y - 6 = 3(x +  2)

5 0
1 year ago
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