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expeople1 [14]
3 years ago
14

Reduce to lowest terms.

Mathematics
2 answers:
fiasKO [112]3 years ago
5 0

Answer:

your friend got that right

Step-by-step explanation:

Furkat [3]3 years ago
3 0
Hi friend,
((5+a)/a)/((a^2 -25) /5a)
=(a+5)/((a^2 - 25)/5)
=(a+5)(5)/(a+5)(a-5)
=5/(a-5)
Therefore your answer is B)
Hope this helps you!
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I need it fasttttttttttttttttttttttttttttttttttttttttttttt<br> pls help
Otrada [13]

Answer:

a. 16 sq cm

b.20 sq cm

Step-by-step explanation:

Area of a parallelogram:

A = b•h

A = 4•4 = 16 sq cm

Area of a trapezoid:

A = 1/2(b+b)•h

A = 1/2(6+4)•4

A = 1/2(10)•4

A = 20 sq cm

6 0
3 years ago
the price of a computer after discount was 1,200 if the doscount was 20% how much discount does mandy get ??
zzz [600]
1200= 80%  
?       =100%

to find the ?  you can multiply 100(1200)=120000
then 120000 divided by 80= 1500
 the original price was 1500, with the discount of  20%   you save 300$   and the price is 1200$

7 0
3 years ago
Read 2 more answers
Find cos θ given that cos 2θ = 5/6 and 0 ≤ θ &lt; π/2. Give an exact answer
trasher [3.6K]
\bf \textit{Double Angle Identities}&#10;\\ \quad \\&#10;sin(2\theta)=2sin(\theta)cos(\theta)&#10;\\ \quad \\&#10;cos(2\theta)=&#10;\begin{cases}&#10;cos^2(\theta)-sin^2(\theta)\\&#10;1-2sin^2(\theta)\\&#10;\boxed{2cos^2(\theta)-1}&#10;\end{cases}&#10;\\ \quad \\&#10;tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\&#10;-------------------------------\\\\&#10;

\bf cos(2\theta)=\cfrac{5}{6}\implies 2cos^2(\theta)-1=\cfrac{5}{6}\implies 2cos^2(\theta)=\cfrac{5}{6}+1&#10;\\\\\\&#10;2cos^2(\theta)=\cfrac{11}{6}\implies cos^2(\theta)=\cfrac{11}{12}\implies cos(\theta)=\pm\sqrt{\cfrac{11}{12}}


now, bear in mind, the square root gives us +/- versions, so, which is it? well, we know the angle is in the range of "<span>0 ≤ θ < π/2", that simply means the 1st quadrant, so, we'll use the positive one then

</span>\bf cos(\theta)=\cfrac{\sqrt{11}}{\sqrt{12}}\implies cos(\theta)=\cfrac{\sqrt{11}}{2\sqrt{3}}&#10;\\\\\\&#10;\textit{now, let's rationalize the denominator}&#10;\\\\\\&#10;\cfrac{\sqrt{11}}{2\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{11}\cdot \sqrt{3}}{2\sqrt{3^2}}\implies \cfrac{\sqrt{11\cdot 33}}{2\cdot 3}\implies \boxed{\cfrac{\sqrt{33}}{6}}<span>
</span>
3 0
4 years ago
Which of the following is the proper name for the figure below?
SOVA2 [1]

Answer:

Option (D)

Step-by-step explanation:

Endpoints of the sides of any polygon are called as vertices. Any polygon is named by its vertices either in a consecutive order either clockwise or counterclockwise.

In the picture attached,

Vertices of the triangle or endpoints of the sides of the polygon are A, T and X.

Therefore, we can name this triangle as ΔATX, ΔTXA, ΔXAT or ΔXTA, ΔAXT, ΔTAX.

Option (D) will be the answer.

7 0
4 years ago
Please help me on this
rewona [7]

Answer:

134

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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