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salantis [7]
4 years ago
13

Circle A is shown on the coordinates plane shown above

Mathematics
1 answer:
omeli [17]4 years ago
7 0

The center of the circle is at (-2, 0). The radius is 5 units. The equation for a circle, in standard form is

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

We have our h and k from the center. h is the x coordinate and k is the y coordinate. The radius is 5. That means that the circle's equation, before simplifying, is

(x-(-2))^2+(y-0)^2=5^2. After simplifying it is

(x+2)^2+y^2=25. That's A and B. For C, we will simply pick points on the axes. One point on the circle is at (3, 0); another point is at (-2, 5); a third point is at (-7, 0). For D, the area of a circle is

A=\pi r^2.

For us that looks like this

A=\pi (5)^2. In terms of pi the answer is 25 pi, but in decimal form to the nearest square unit, it's 79 units squared.

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hram777 [196]
The area of the triangle is 
a=( m x n) /2  in cm² , implies   2xa = m xn
the area of the rectanglle
A= n x m  in cm²
but we know  <span>  2xa = m xn = A,  </span>
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7 0
4 years ago
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(x-5)1/2+5=2<br> (x-5)1/2=-3<br> [(x-5)^1/2]^2=(-3)^2
Studentka2010 [4]

Answer:

x = 14

Step-by-step explanation:

You want to solve for x, right?

(x-5)^(1/2)+5=2

You have it right so far.

(x - 5)^(1/2) = -3

(x - 5)  =  (-3)^2

x - 5 = 9

x = 9 + 5 = 14

3 0
4 years ago
Is the area of ABC round to the nearest tenth
svlad2 [7]
Hi there!

Since we know that this is a right isosceles triangle, we can use the 45-45-90 relationship to figure out the missing side lengths. Using this relationship, we know that the two marked sides are both equal to 5 inches. Next, we need to use the formula to find the area of a triangle, which is A = 1/2(b * h). We know that the base of this triangle is 5, as is the height. All we need to do is plug in the numbers and solve: A = 1/2(5 * 5) | A = 1/2(25) | A = 12.5 inches^2. 

Hope this helps!! :)
If something doesn't make sense or if there's anything else that I can help you with, please let me know!
5 0
3 years ago
The line tangent to the graph of g(x) = x^{3}-4x+1 at the point (2, 1) is given by the formula
Usimov [2.4K]

well, about A and D, I just plugged the values on the slope formula of

\bf \begin{array}{llll} g(x)=x^3-4x+1\\ L(x) = 8(x-2)+1 \end{array} \qquad \begin{cases} x_1=1.9\\ x_2=2.1 \end{cases}\implies \cfrac{f(b)-f(a)}{b-a}

for A the values are 8.01 and 8.0, so indeed those "slopes" are close. \textit{\huge \checkmark}

for D the values are -2.25 and 8.0, so no dice on that one.

for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.

checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice. \textit{\huge \checkmark}

for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.

for E, do they intercept at (2,1)?  well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent. \textit{\huge \checkmark}

for F, we know the slope of the line L(x) is 8, is g'(2) = 8?  let's check

recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice. \textit{\huge \checkmark}

6 0
4 years ago
|a − b + c| , if a=−8; b=−5; c=1<br> please help thx brosuf
Arisa [49]

Answer:

2

Step-by-step explanation:

|a - b + c|  =  | - 8 + 5 + 1|  =  | -  2|  = 2

8 0
3 years ago
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