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Nitella [24]
3 years ago
15

Does anyone know any other good apps for math like Algebra 1.?

Mathematics
2 answers:
vesna_86 [32]3 years ago
8 0
Malmath is really good app too and it works offline.
Novay_Z [31]3 years ago
7 0
Photomath is a good one. So is KhanAcademy, but I don't know if they have an app version of the website.
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F(x)=-5x+2 and g(x)=1/2x+4 find f(g(12))
skad [1K]

Answer:

- 48

Step-by-step explanation:

If you want to learn more about this concept, it's called composition of functions.

First, you plug in g(x) as if it were x into f(x).

f(g(x))= -5 (1/2x + 4) + 2

= -5/2x - 20 + 2

= -5/2x - 18

Then, plug in the value given, as x.

= -5/2 (12) - 18

= -30 - 18

= - 48

I hope this helped!

3 0
3 years ago
What is the awnser to this question because I don't understand
Virty [35]
Pemdas 1. parentheses: everything is already in parentheses. so next is exponents. 10×-2 is -20. and 10×6 is 60. so 4×-20 is the next step. it equals -80. and then 3×60 is 180. lastly,add -80 and 180 together. the answer should be 100 if i did that correctly.
4 0
3 years ago
HELP PLZ RN&lt;3<br> which graph displays a rate of change of 2.5?
Ilya [14]

Answer:

The second graph.

Step-by-step explanation:

The first graph has a negative slope, so that can't be the right answer.

The last graph isn't steep enough to show a rate of 2.5.

This leaves us with the middle graph which is correct.

3 0
3 years ago
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha
mr Goodwill [35]

Answer:

Step-by-step explanation:

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by

T(x) = 160-0.05x^2

a. [0, 10]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

The average temperature

= (160 + 155)/2 = 157.5

b. [10, 40]

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (80 + 155)/2 = 117.5

c. [0, 40]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

6 0
2 years ago
V-3/-10= -4 please answer step by step
Triss [41]

See the pic above!

Hope it helps!

5 0
3 years ago
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