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3 years ago

11

please help will give BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST BRAINLI

EST BRAINLIEST BRAINLIEST for correct answer!!!!!!!!!!!!!

2 answers:

o-na [289]3 years ago

8 0

Answer:

25 inches is the answer

Temka [501]3 years ago

3 0

Answer:

25 in or 20in

Step-by-step explanation:

it is not half way if it was it would be 15in its way of half way

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Written as a simplified polynomial in standard form, what is the result when

Naddika [18.5K]

Step-by-step explanation:

7x+10 - (x+2)^2

7x+10 - (x^2 + 2x + 4)

7x+ 10 - x^2 - 2x - 4

5x + 6 - x^2

then if u have to find the value of x then put the value of a = 1,b = 5,c = 6 in quadratic formula (-b + under root b^2 - 4ac / 2a) or (- b - under root b^2 - 4ac / 2a) it will gives the value of x...

HOPE it u..!!

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7 0

3 years ago

Select the correct answer.

ella [17]

B it’s b I think because it’s b I’m pretty sure b I’m also not sure b

3 0

3 years ago

Match the two step inequality to the correct graph.

miskamm [114]

Answer:

Hello,

I think -2p < - 10 goes with graph one

Then -M = 8 > 6 goes to Graph Two

Next 13 > 4r + 5 goes to Graph three

Last but not least -7 - Z < -1 goes to Graph Four

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Step-by-step explanation:

3 0

3 years ago

Determine whether angle ABC should be solved by using the Law of Sines or the Law of Cosines. Then solve the triangle.

drek231 [11]

Answer:

Law of Cosines

Angle A = 45°

Angle B = 51°

Angle C = 84°

Step-by-step explanation:

Law of sines is used when we are given

a) two angles and one side or

b) two sides and non-included angle

Law of cosines is used when we are given

a) three sides or

b) two sides and included angle

In the given question we are given three sides so, Law of Cosines will be used to solve the triangle.

Law of Cosines is:

$a^2 = b^2 + c^2 -2bccos A\\b^2 = a^2 + c^2 -2accos B\\c^2 = a^2 + b^2 -2abcosC$

We will find the three angles A ,B and C of the triangle using above formula.

a= 10, b=11, c=14

Putting values and finding angle A

$a^2 = b^2 + c^2 -2bccos A\\(10)^2 = (11)^2 + (14)^2 -2(11)(14)cosA\\100 = 121 + 196 -308cosA\\100 = 317 -308 cosA\\100-317 = -308cosA\\-217/-308 = cos A\\0.704 = cos A\\=> A = cos ^{-1}(0.704)\\A= 45$

Now finding angle B

$b^2 = a^2 + c^2 -2ac cos B\\(11)^2 = (10)^2 + (14)^2 - 2(10)(14)cosB\\121 = 100+196 - 280cosB\\121 -296 = -280cosB\\-175/-280 = cosB\\0.625 = cosB\\=> B = cos^{-1} (0.625)\\B = 51$

Now finding angle C

$c^2 = a^2 + b^2 -2abcosC\\(14)^2 =(10)^2 + (11)^2 -2(10)(11)cosC\\196 = 100+121 -220cosC\\196 -221 = -220cosC\\-25/-220 = cos C\\0.11 = cosC\\=> C = cos^{-1}(0.11)\\C= 84$

4 0

3 years ago

Read 2 more answers

Can someone try doing this plz

Svetach [21]

Answer:

F

Step-by-step explanation:

Your welcome my guy :D

5 0

3 years ago

Read 2 more answers