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3 years ago

HELPPP WIILLLL MAKR BRAINLIEST IF RIGHTTTT

Mathematics

2 answers:

goldenfox [79]3 years ago

8 0

Answer:

4π yd²

Step-by-step explanation:

Area of a circle = πr²

πr²=4π

r²=4

r=2

Circumference of a circle = 2πr = 2π(2) =4π

kondor19780726 [428]3 years ago

7 0

Answer:

I think the answer is letter B

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Peter babysits his sister on Monday and Friday. He babysits for his neighbor 40% of the days Monday through Friday. He babysits

saul85 [17]

Answer:

Question 1: False

Question 2: True

Question 3: True

Question 4: False

Step-by-step explanation:

Peter babysits his sister 2 days: Monday and Friday.

He babysits for his neighbor 40% of the time between Monday through Friday. This includes 5 days. 40% of 5days = 2 days.

On the weekend, Peter babysits 50% of the days. There are 2 days in the weekend

50% × 2 days = 1 day.

Question 1: False

Altogether, Peter babysits 5 days every week.

Question 2: True

Question 3: True

Peter babysits 4 out 5 days during Monday-Friday 4/5= 80%

Question 4: False

Peter babysits 5 days out of 7 during the week. 5/7 = 71% of the days, which is less than 90%

5 0

4 years ago

Write an equation of a circle with a center at (2, - 4) and a radius of 4.

fgiga [73]

9514 1404 393

Answer:

(x -2)^2 +(y +4)^2 = 16

Step-by-step explanation:

The equation of a circle with center (h, k) and radius r is ...

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

You have (h, k) = (2, -4) and r = 4, so your equation is ...

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

(x -2)^2 +(y +4)^2 = 16

6 0

3 years ago

There are 6 blades on each windmill how many total blades are on 7 vwindmills? use fives facts to solves

aliina [53]

6 times 7 = 42 hope helped

8 0

3 years ago

The graph of which function has an axis of symmetry at x =-1/4 ?

Phoenix [80]

we know that

The equation of the vertical parabola in vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

$x=h$ ------> axis of symmetry of a vertical parabola

we will determine in each case the axis of symmetry to determine the solution

case A) $f(x)=2x^{2}+x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=2x^{2}+x$

Factor the leading coefficient

$f(x)+1=2(x^{2}+0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+0.125=2(x^{2}+0.5x+0.0625)$

$f(x)+1.125=2(x^{2}+0.5x+0.0625)$

Rewrite as perfect squares

$f(x)+1.125=2(x+0.25)^{2}$

$f(x)=2(x+0.25)^{2}-1.125$

the vertex is the point $(-0.25,-1.125)$

the axis of symmetry is

$x=-0.25=-\frac{1}{4}$

therefore

the function $f(x)=2x^{2}+x-1$ has an axis of symmetry at $x=-\frac{1}{4}$

case B) $f(x)=2x^{2}-x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=2x^{2}-x$

Factor the leading coefficient

$f(x)-1=2(x^{2}-0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+0.125=2(x^{2}-0.5x+0.0625)$

$f(x)-0.875=2(x^{2}-0.5x+0.0625)$

Rewrite as perfect squares

$f(x)-0.875=2(x-0.25)^{2}$

$f(x)=2(x-0.25)^{2}+0.875$

the vertex is the point $(0.25,0.875)$

the axis of symmetry is

$x=0.25=\frac{1}{4}$

therefore

the function $f(x)=2x^{2}-x+1$ does not have a symmetry axis in $x=-\frac{1}{4}$

case C) $f(x)=x^{2}+2x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=x^{2}+2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+1=x^{2}+2x+1$

$f(x)+2=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+2=(x+1)^{2}$

$f(x)=(x+1)^{2}-2$

the vertex is the point $(-1,-2)$

the axis of symmetry is

$x=-1$

therefore

the function $f(x)=x^{2}+2x-1$ does not have a symmetry axis in $x=-\frac{1}{4}$

case D) $f(x)=x^{2}-2x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=x^{2}-2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+1=x^{2}-2x+1$

$f(x)=x^{2}-2x+1$

Rewrite as perfect squares

$f(x)=(x-1)^{2}$

the vertex is the point $(1,0)$

the axis of symmetry is

$x=1$

therefore

the function $f(x)=x^{2}-2x+1$ does not have a symmetry axis in $x=-\frac{1}{4}$

the answer is

$f(x)=2x^{2}+x-1$

7 0

4 years ago

Read 2 more answers

I’ll give you brainliest , pls help

NeX [460]

Answer:

your answer is correct

Step-by-step explanation:

7 0

3 years ago