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Cerrena [4.2K]
3 years ago
9

Please help me with 11th grade math // Geometry ​

Mathematics
1 answer:
ohaa [14]3 years ago
5 0

Answer:

I sent I the answers on ur account

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Rationalize the denominator and simplify <br> √a+1-2/√a+1+2
zalisa [80]

5+a-4√(a+1)/(a-3) is the simplified form of the number. This can be obtained by multiplying numerator and denominator with conjugate of the denominator.

<h3>Simplify the number:</h3>

Given the number,

√a+1-2/√a+1+2

Denominator = √a+1+2

Conjugate of the denominator = √a+1 - 2

Now, (√a+1-2)(√a+1 - 2 )/√(a+1+2)(√a+1 - 2 )

=(√a+1 - 2 )²/(√a+1)² - 4

=(a+1 -4√a+1+5)/(a+1-4)

=5+a-4√(a+1)/(a-3)

Hence 5+a-4√(a+1)/(a-3) is the simplified form of the number.

Learn more about rational numbers:

brainly.com/question/1310146

#SPJ1

7 0
2 years ago
10. Find the value of x.<br> (5x+12)° (3x+8)°<br> 10<br> 15<br> 20<br> 25
satela [25.4K]

Answer:

3x+8+5x+12

8x+20=180

A. 80+20=100 No

B. 120+20=140 No

C. 160+20=180 Yes

D. 200+20=220 No

The Answer is C

Hope This Helps!!!

3 0
3 years ago
Read 2 more answers
If you take a 100 dollars from register then use the money to buy 70.00 purchase then get back 30.00 how much money is lost .
Anton [14]
You would do 100-30 because 30 is what you have left and 100-30 is 70 so $70 was lost
3 0
4 years ago
On highway 77, the Dobson exit is halfway between the Mt. Airy exit and the Elkin exit. If the Elkin exit is located at (-5,-2)
Leya [2.2K]

Given :

The Dobson exit is halfway between the Mt. Airy exit and the Elkin exit. If the Elkin exit is located at (-5,-2) and the Dobson exit is located at (-1,4).

To Find :

How far is it from the Elkin exit to the Mt. Airy exit.

Solution :

E(-5,-2) , D(-1 ,4 ) .

It is also given that Dobson exit is halfway between the Mt. Airy exit and the Elkin exit.

Let , location of Elkin exit is A(h,k) .

So , point D in terms of A and E is given by :

(\dfrac{h-5}{2},\dfrac{k-2}{2}) .

Comparing it with given D :

\dfrac{h-5}{2}=-1\\\\h=3 \dfrac{k-2}{2}=4\\\\k=10

Therefore , the location of Mt. Airy exit is ( 3, 10 ) .

Distance between Elkin exit to the Mt. Airy exit.

D=\sqrt{(5-3)^2+(-2-10)^2}\\\\D=12.17\ units

Therefore , distance between Elkin exit to the Mt. Airy exit is 12.17 units .

Hence ,this is the required solution .

4 0
3 years ago
ÐABC = 28°, and ÐBCA is larger than ÐABC. Triangle ABC is an isosceles triangle. What is the measure of the other two angles?
barxatty [35]
ÐABC = 28°, and ÐBCA is larger than ÐABC. Triangle ABC is an isosceles triangle. It is stated that in an isosceles triangle the two angles are congruent to each other and since one of the angle is given as 28° the measures of two remaining angles is 76°.
7 0
4 years ago
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