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3 years ago

Wanda shops for books during a 20% off sale.

Mathematics

2 answers:

olga55 [171]3 years ago

8 0

Answer: mark brainliest im grinding ranks

Step-by-step explanation:

babymother [125]3 years ago

7 0

Answer:

I am confused can you explain a little more

Step-by-step explanation:

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PLEASE HELP! While shopping, Kyla found a dress that she would like to purchase, but it costs $52.25 more than she has. Kyla cha

zhenek [66]

I don’t have a double number line but she would have to spend 9.5 (9 1/2) hours babysitting to earn enough money to buy the dress

7 0

3 years ago

Find the value of x 90, 68 2x

Ivan

Answer:

2x + x +90= 180 We will add 2x + x=3x We get 3x + 90 =180 Now we subtract 3x = 180–90 ... x=30, for confirmation put value of x =30in equation and verify. LHS=RHS. Thats it. 68 views.

8 0

3 years ago

What is the answer for question 1 and how

quester [9]

Step-by-step explanation:

ABC is an isoceles triangle (both legs are equally long). and AB is its baseline.

OC is now a median for ABC splitting the angle at C and AB in half.

so, we have 2 right-angled triangles : OAC and OBC.

the half-angles at C are 42/2 = 21°.

the angles at A and B are 90°.

and the half-angles at O are 180 - 90 - 21 = 69°.

remember that the sum of all angles in a triangle must always be 180°.

AB are the heights of both of these triangles.

the single height is sin(69)×7 = 6.535062985... cm

and so,

AB = 2× height = 13.07012597... cm.

4 0

2 years ago

Read 2 more answers

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers

How can you use rates to compare the cost of two cereal boxes that are different sizes.

Anarel [89]

You subtract and devide i beleve. hope iu helped sorry if wrong.

8 0

3 years ago