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Citrus2011 [14]
3 years ago
12

If the outside temperature was 72 yesterday and today is 16 degrees cooler what is today's temperature

Mathematics
2 answers:
Advocard [28]3 years ago
6 0
The answer is 56 degrees since 72-16 is 56
Natalija [7]3 years ago
5 0

Answer:

56 degrees Fahrenheit

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Ania scored a total of 120 points this season so far and averaged 15 points a game. How many games has she played?​
arlik [135]
She played 8 games. Since she earned a total of 120 points and 15 points per game you would divide 120/15=8
4 0
2 years ago
Hi can someone check my answers please? thanks, :).
gregori [183]
The 3rd answer is wrong, the answer is 19.
The 5th answer is wrong, the answer is -9.
The 7th answer is wrong, the answer is 8.
The 11th answer is wrong, the answer is 2.
The 12th answer is wrong, the answer is 10.

Hope I could help :)
5 0
3 years ago
Read 2 more answers
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
Suppose f(x)=x^2+3 and g(x)=x-2 find (f-g)(x
SSSSS [86.1K]

Answer:

x^2-x+5

Step-by-step explanation:

(f-g)(x)=f(x)-g(x)

= (x^2 + 3) - (x - 2)   (Don’t forget the -1 multiplies to the x AND -2

= x^2 + 3 - x + 2

=x^2-x+5

6 0
3 years ago
Can you help me solve? Question is on top left
galina1969 [7]

Answer:

See below ~

Step-by-step explanation:

⇒ 65 = 65 + y (alternate exterior angles)

⇒ y = 0

⇒ 4(12x + 1) + 15 = 180 - 65 - y (Linear angles)

⇒ 48x + 4 = 100 - 0

⇒ 48x = 96

⇒ x = 2

⇒ 3(7z + 6) + 5(5z + 1) = 115 (Both the angles are equal "alt. ext. angles")

⇒ 21z + 18 + 25z + 5 = 115

⇒ 46z + 23 = 115

⇒ 46z = 92

⇒ z = 2

3 0
2 years ago
Read 2 more answers
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