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BlackZzzverrR [31]
3 years ago
6

PLS HELP! BAD AT MATH!

Mathematics
1 answer:
Alexandra [31]3 years ago
5 0

Answer:

umm you may need to try Photo Math

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PLease answer as soon as possible
yulyashka [42]

Answer:

No

Step-by-step explanation:

Every x value in the domain should only have 1 y value. In the graph, for x=5 (the input) there are 2 y values (6 and -6) so this is not a function.

3 0
3 years ago
Read 2 more answers
You need to cross a canal and want to determine the distance across the opposite side. Since you're able to take measurements on
MariettaO [177]

Answer:

36.7 ft

Step-by-step explanation:

Measurements for the sides of the canal is given as: 40 ft. and 16 ft.

You solve the above question using Pythagoras Theorem

The distance across the canal is calculated as:

√(40² - 16²)

= √(1344)

= 36.66060556 ft

Approximately = 36.7 ft

Therefore, the distance x across the canal = 36.7 ft

8 0
3 years ago
Find f(x) and g(x) so that the function can be described as y = f(g(x)). <br><br> y = 7/sqrt 8x +10
Blizzard [7]
Assuming you meant

y =  \frac{7}{ \sqrt{8x+10} }

Then you can compose the function g(x)

g(x)= \sqrt{8x+10}

With the function

f(x)= \frac{7}{x}


Because f (g(x) ) means that you first find (calculte)     \sqrt{8x+10}       and second you divide 7 by g(x) to obtain 7 / [√(8x+10) ].
8 0
3 years ago
Evaluate using <br> Definite integrals
swat32

Since [0,4]=[0,1]\cup(1,4], we can rewrite the integral as

\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt

Both integrals are quite immediate: you only need to use the power rule

\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}

to get

\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4

Now we only need to evaluate the antiderivatives:

\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15

So, the final answer is 15.

4 0
3 years ago
Round 2,795 to the nearest ten
Ahat [919]

Answer:

2,800

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
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