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kati45 [8]
2 years ago
9

Which of the following is equivalent to 7a^4+3a^4

Mathematics
2 answers:
DanielleElmas [232]2 years ago
8 0

Answer:

Step-by-step explanation:

7a^4 + 3a^4 is 10a^4

nasty-shy [4]2 years ago
3 0
Give me a list for me to answer

2,401a+81a
is equivalent
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Can some help me out with this EXTRA POINTS!!
Butoxors [25]

Answer: ≈ 185.73009

Step-by-step explanation:

area of square: length^{2}

area of circle: \frac{radius^{2} * \pi }{y}

length: 15

radius = diameter/2

radius: 5

area of square - area of circle =  area between square and circle:

15^{2} - \frac{5^{2} * \pi }{2} = 225 - \frac{25\pi}{2} ≈ 185.73009

hope it helps!

Please mark as brainliest.

8 0
2 years ago
Locate the points of discontinuity in the piecewise function shown below.
algol13

Answer:

Step-by-step explanation:

The given piecewise function i

From the given function it is clear that function is divided at x=-1 and x=2. It means we check the discontinuity at x=-1 and x=2.

For x=-1,

LHL:  

Since LHL ≠ f(-1), therefore the given function is discontinuous at x=-1.

For x=2,

LHL:  

Since LHL ≠ f(2), therefore the given function is discontinuous at x=2.

Therefore, the correct option is A.

8 0
3 years ago
What is the x-intercept of the equation 3x - 2y = -12 ​
muminat
X=2/3y-4

EXPLANATION
3x-2y=-12
You take -2y to the other side
It becomes 3x=2y-12
Then you divide 3
X=2/3y
-12/3=-4
X=2/3y-4
4 0
3 years ago
9 x 76 using the distributive property
zysi [14]

As far as I'm aware, you can't use the distributive property for a simple multiplication problem.  However, the answer to 9 x 76 is 684.  Hope this helped!

-TTL

5 0
3 years ago
Read 2 more answers
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
3 years ago
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