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attashe74 [19]
3 years ago
8

Solve: 3x + 18 - 7 = 41

Mathematics
1 answer:
solmaris [256]3 years ago
7 0

Answer:

x + 10

Step-by-step explanation:

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Determine the equation of the circle graphed below.
Wittaler [7]

Answer:

(x - 6)^2 + (y + 4)^2 = 4

Step-by-step explanation:

formula for a circle

( x - h )^2 + ( y - k )^2 = r^2

(h,k) is the center

r is the radius

(6,-4) is center

2 is radius

Substitute in numbers and watch the negatives.

6 0
3 years ago
Is the function y=0.5x linear?
Dmitry [639]
Yes.

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be
0
0
or
1
1
for each of its variables. In this case, the degree of variable
y
y
is
1
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and the degree of variable
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4 0
3 years ago
Read 2 more answers
Will mark brainliest if answered correctly!
Dima020 [189]
<h3>Answer: </h3>

about 1.768 seconds

<h3>Explanation:</h3>

Since the phone is <em>dropped</em>, the first equation applies. The final height is assumed to be zero, so we have ...

... h(t) = 0 = -16t² +50

... 16t² = 50 . . . . . . . . add 16t²

... t² = 50/16 . . . . . . . . divide by 16

... t = √3.125 . . . . . . . take the square root

... t ≈ 1.768 . . . . . . . . round to milliseconds

5 0
3 years ago
Can someone please help me with question 35?
Genrish500 [490]
First you would find the area of the triangle which would be 84×92 then divide that answer by 2.
Then you will find the area of the circle using the given radius 20.5.
Lets work it out.
84×92=7728
7728÷2=3864 ft²
3864 ft² is the area of the backyard including the pool.
A=pi(20.5)²
A=1320.25 ft²
1320.25 ft² is the area of the pool.
Now we have to subtract to get the area of the backyard without the pool.
3864-1320.25=2543.75 ft²
2543.75 ft² is your answer.
Hope This Helps and God Bless!
8 0
3 years ago
The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperatu
PSYCHO15rus [73]

Answer:

a) I(95,50) = 73.19 degrees

b) I_{T}(95,50) = -7.73

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

a) Calculate I at (T ,H) = (95, 50).

I(95,50) = 45.33 + 0.6845*(95) + 5.758*(50) - 0.00365*(95)^{2} - 0.1565*95*50 + 0.001*50*95^{2} = 73.19 degrees

(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is I_{T}(T,H). So

I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}

I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H

I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73

8 0
3 years ago
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