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DaniilM [7]
3 years ago
15

Solve the equation -14+2b+= -2b+2​

Mathematics
2 answers:
disa [49]3 years ago
8 0

Answer:

b=4

Step-by-step explanation:

-14+2b= -2b+2​

Add 2b to each side

-14+2b+2b= -2b+2b+2

-14+4b =2

Add 14 to each side

-14+4b+14= 14+2​​

4b = 16

Divide each side by 4

4b/4 = 16/4

b =4

Harrizon [31]3 years ago
3 0
-14+2b=-2b+2

2b-14+2b=-2b+2+2b
4b-14=2

4b-14+14=2+14
4b=16

4b/4 = 16/4
B= 4

ANSWER:
b=4
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Natasha2012 [34]

The area of the given circle is 95 square ft.

Calculation for the Area of the Circle:

It is given in the diagram that,

The diameter of the circle, d = 11 ft.

Now, we know that the radius of a circle is equal to half of the diameter.

Therefore, radius of the given circle, r = 11/2 ft.

The formula for the area of the circle is given as follows,

A = π × r²

Substituting the values, π = 3.1, and r = 11/2, we get,

A = (3.14) × (11/2)²

A =  (3.14) × 30.25

A = 94.985

A ≈ 95

Hence, the area of the given circle with diameter 11 ft. comes out to be 95 ft².

Learn more about a circle here:

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7 0
2 years ago
Read 2 more answers
15
aleksandr82 [10.1K]

NO THOS IS NOT TRUESNBDJDN

8 0
3 years ago
Here's the deal. I give you a question, you give me an answer. I give you points and Brainliest if possible. You use the points
olga55 [171]

Answer:

f(x) = -(x-3)^2 + 4

Step-by-step explanation:

(h; k) are the coordinates of the vertex.

On the graph are (3, 4), therefore we have:

f(x) = a(x-3)^2 + 4

We have the x- intercept (1,0) -> x=1; y=0.

Substitute them into the equation:

0 = a(1 - 3)^2 + 4

0 = a(-2)^2 + 4

4a + 4 = 0 | -4

4a = -4   |-4

a = -1

So, we have the answer:

f(x) = -(x-3)^2 + 4

5 0
3 years ago
Slope=10,y-intercept = - 5
marin [14]

Answer:

the equation of the above is

y= 10x - 5

7 0
2 years ago
The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet
vladimir1956 [14]
\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}

\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
-------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}
8 0
3 years ago
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