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2 years ago

Add (4x^2+x-9) and (3x^2+x+4)

Mathematics

1 answer:

HACTEHA [7]2 years ago

7 0

Answer:

7x^2 + 2x - 5

Step-by-step explanation:

(4x^2 + x - 9) + (3x^2 + x + 4)

7x^2 + 2x - 5

Pretty simple, just add 4 and 3, then x (or 1) and x, lastly -9 and 4. Then you get your final equation.

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The volume of a cube is 216 cubic centimeters. How long is each side?

Zarrin [17]

Answer:

Step-by-step explanation:

3 0

2 years ago

Find the value of x and the value of y

serg [7]

Answer:

Step-by-step explanation:

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

use Cos for y .... b/c we see that function above has all the pieces we are going to work with

use Sin for x .. for the same reason above

Cos(30) = y / 4 $\sqrt{2}$

4 $\sqrt{2}$ Cos(30) = y

4 $\sqrt{2}$ $\sqrt{3}$ /2 = y

2 $\sqrt{6}$ = y

leave the answer for y like that.. it's in 'exact" form

now for x

Sin(30) = x / 4 $\sqrt{2}$

4 $\sqrt{2}$ Sin(30) = x

4 $\sqrt{2}$ * 1/2 = x

2 $\sqrt{2}$ = x

and leave your answer for x like that as it's in it's exact form. :)

oh and I see the answer in the list.. do you? :)

8 0

2 years ago

A wire that is 22 feet long connects the top of a pole to the ground. The wire is attached to the ground at a point that is 10 f

krok68 [10]

✰ Concept Used :-

⠀

In this question, we can clearly observer that the diagram shows a right angled triangle. And, we have been provided with the value of base, and the value of hypotenuse, using the pythagoras theorem, now we can easily find out the value of the perpendicular i.e. the value of the side h. According to the pythagoras theorem, square of hypotenuse is equal to the sum of square of perpendicular and square of side respectively. Therefore, square of side is equal to the difference of square of hypotenuse and square of perpendicular.

⠀

✰ Given Information :-

⠀

Hypotenuse = 22 ft.
Base = 10 ft.

⠀

✰ To Find :-

⠀

The value of side or the perpendicular

⠀

✰ Formula Used :-

⠀

$\star \: \underline{ \boxed{ \purple { \sf {Side}^{2} = {Hypotenuse}^{2} - {Base}^{2} }}} \: \star$

⠀

✰ Solution :-

⠀

$\sf \longrightarrow {Side}^{2} = {(22 \: ft)}^{2} - {(10 \: ft)}^{2} \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{2} = {484 \: ft}^{2} - {100 \: ft}^{2} \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{2} = {384 \: ft}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{} = \sqrt{ {384 \: ft}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{} = \underline{ \boxed{ \frak{ \green{19.60 \: ft}}}} \: \star \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\$