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3 years ago

Do I multiply, area, explain pkease

Mathematics

1 answer:

KIM [24]3 years ago

8 0

Step-by-step explanation:

Area of a trapezoid is 1/2h(b1+b2)

Height is 8 in. Base 1 is 5 in and Base 2 is 10 in. Do not use the diagonal side.

For the triangle-scalene triangle area formula is bh/2 (base times height divided by 2) The base is 5 in and the height is 4 inches.

For the paralellogram multiply base by height. 6 multiplied by 5 is 30

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nora is planning a birthday party for her little sister, colleen. she needs to purchase 12 cupcakes she can not spend more than

Lapatulllka [165]

Answer:

if she cannot spend more than $40, she will spend .3 on each cupcake

Step-by-step explanation:

when you divide 12 by 40, you get .3 and to check if it is true, you can multiply .3 x 40 and you will get 12 hope you get it right, and have a good day :)

6 0

2 years ago

Help please bad please

Sergio [31]

Answer:

$\displaystyle A .20$

Step-by-step explanation:

By tangent theorem we acquire:

$\displaystyle x + 10 = 9 + 21$

simplify addition:

$\displaystyle x + 10 = 30$

cancel 10 from both sides:

$\displaystyle x = 20$

hence,

our answer is A

4 0

2 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

I NEED HELP PLEASE ASAP!!! :)

erik [133]

Answer:

r = 6.4/(1+sin(θ))

Step-by-step explanation:

As the attachment shows, for the given directrix and eccentricity, the equation is ...

$r=\dfrac{ed}{1+e\sin{\theta}}\\\\r=\dfrac{1.6\cdot 4}{1+1.6\sin{\theta}}\\\\\boxed{r=\dfrac{6.4}{1+\sin{\theta}}}\qquad\text{matches the 2$^{\text{nd}}$ choice}$