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2 years ago

13

Who can help me with this algebra thanks for your help guys.

1 answer:

timama [110]2 years ago

8 0

Answer:

30. 25÷5 = 10÷2

31. 3×3 = 6+3

33. 13+19 = 8×4

34. 14-2 = 6×2

Step-by-step explanation:

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What are the solutions to the equation 0 = x2 – x – 6?

ioda

Factor to this equation out, the easiest way is to solve for the two x values that add up to -1 but when multiplied is equal to -6. It should look like this: (x +- ?)(x +- ?).
The factored out form is then (x - 3)(x +2), because -3 + 2 = -1 and when multiplied together gives -6.
x - 3 = 0
x = 3
x + 2 = 0
x = -2
The two solutions are therefore, x = 3 and x = -2.

Hope it helps

3 0

2 years ago

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What is the square root of y^3

tia_tia [17]

Answer:

$y^{\frac{3}{2}$

Step-by-step explanation:

Let $m=\sqrt{y^3}$

$m=(y^3)^{\frac{1}{2}}\\\\m=y^{3\times \frac{1}{2}}\ \ \ \ \ \ \ \ \ [as\ (x^a)^b=x^{ab}]\\\\m=y^{\frac{3}{2}$

7 0

2 years ago

is it possible to draw two congruent triangles so that one triangle is an acute triangle and one triangle is right triangle? why

BaLLatris [955]

No because then they would be different triangles. Congruent triangles are triangles that are the same

I hope I've helped!

4 0

2 years ago

What is 1×3+45÷1×0=

QveST [7]

The answer to this equation would be 3.

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3 years ago

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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago