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3 years ago

What are the two square roots of 121/49

Mathematics

2 answers:

marusya05 [52]3 years ago

6 0

Answer:

11/7, I copied it from the other answer, and I need brainiest if i'm going to rank up before I hit triple the points needed for my rank-up.

sasho [114]3 years ago

5 0

11/7, Hope this helps.

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Can some one help me and solve this problems

Lubov Fominskaja [6]

Answer:

2 i think i dint sleep last night sorry if im wrong

Step-by-step explanation:

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3 years ago

-11=61-6q what is q solve the equation

aleksley [76]

Answer:

q=12

Step-by-step explanation:

-11=61-6q

-11-61 =-6q

-72=-6q

q=72/6

q=12

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3 years ago

What type of triangle is shown in the image?

frutty [35]

The triangle is an Obtuse triangle

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3 years ago

Read 2 more answers

Is x = 6 a solution to the equation 5(x – 3) = x + 13?

deff fn [24]

Answer:

x=6 is not a solution

Step-by-step explanation:

We can substitute 6 as x to find out:

5(6-3)=6+13

first, because of PEMDAS we'll do 6-3

5(3)=6+13

multiply 5 by 3 and add 6 and 13

15=19

Since the statement is not true, that means that x=6 is not a solution

Hope this helps!

8 0

3 years ago

Convert x2 + y2 -4x = 0 to polar form.

Bingel [31]

Answer:

$r(r-4\cos \theta)=0$

Step-by-step explanation:

We are given the following equation:

$x^2 + y^2 -4x = 0$

We have to convert it into polar form.

We put

$x = r \cos \theta\\y = r\sin \theta$

Putting values, we get:

$x^2 + y^2 -4x = 0\\(r\cos \theta)^2 + (r\sin \theta)^2 - 4(r\cos \theta) = 0\\r^2(\cos^2 \theta + \sin^2 \theta) - 4r\cos \theta = 0\\r^2 - 4r\cos \theta = 0\\r(r-4\cos \theta)=0$

is the required polar form.

7 0

3 years ago