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Harlamova29_29 [7]
3 years ago
11

The radius of a birthday cake is 5 inches. Icing will decorate the top edge of the cake.

Mathematics
1 answer:
maria [59]3 years ago
4 0

Answer:

Step-by-step explanation:

circumference= pi * diameter

diameter= 2 * radius

circumference= pi *2(5)

c= 10pi

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If x+y=5,x-y=4,find the value of x²-y²​
inysia [295]

Answer:

20

Step-by-step explanation:

→ First find the value of x and y

x + y = 5

x - y = 4

→ Add the equations to cancel out the y's

2x = 9

→ Divide both sides by 2 to find the value of x

x = 4.5

→ Substitute x = 4.5 back into x - y = 4 to find the value of y

4.5 - y = 4

→ Minus 4.5 from both sides to isolate -y

-y = -0.5

→ Multiply everything by -1

y = 0.5

→ Substitute x = 4.5 and y = 0.5 into x² - y²

4.5² - 0.5² = 20.25 - 0.25 = 20

4 0
3 years ago
√225x^7y^9
igor_vitrenko [27]

Answer:

√(225x^7y^9) = 15 sqrt(x^7 y^9)

Step-by-step explanation:

4 0
3 years ago
The solution of a+b>−4 is a>−9. What is the value of b?
Lynna [10]

Answer:

b is greater than or equal to 5

5 0
2 years ago
f(x) = x2. What is g(x)?SIf(x)= x/g(x)(2,1)-5O A.06)-()O B. g(x) = 2x2O C. g(x)-9(x)=(*)O D. g(x)= x2
Andrew [12]
Explanation

As you can see all the possible answers have the same form:

g(x)=ax^2

By looking at the picture you'll notice that the graph of g(x) has to pass through the point (2,1). Remember that the points in the graph of g(x) have the form (x,g(x)). Since (2,1) is part of the graph of g(x) then we have the following:

\begin{gathered} (2,1)=(x,g(x)) \\ x=2 \\ g(2)=1 \end{gathered}

So let's evaluate the expression for g(x) that we wrote before at x=2. This way we'll obtain an equation for the number a:

\begin{gathered} g(2)=1=a\cdot2^2 \\ 1=4a \end{gathered}

Then we can divide both sides by 4:

\begin{gathered} \frac{1}{4}=\frac{4a}{4} \\ a=\frac{1}{4} \end{gathered}

Then we get:

g(x)=\frac{1}{4}x^2=(\frac{1}{2}x)^2Answer

Then the answer is option A.

3 0
1 year ago
Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)
marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}

8 0
2 years ago
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