I need to know what 7 x 4 is

Mathematics

1 answer:

Vesna [10]3 years ago

6 0

7 X 4 = 28 if it’s hard for you just add 7, 4 times

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Which expression has exactly three terms? Select all the expressions that apply.

irga5000 [103]

The answer is expressions D, E, and G.

In algebra, a ‘term’ usually means the different parts of an expression that are separated by + and - signs.

Options A and B only have 1 term, an x or y³, so these are incorrect.

Option C has 1 term as well, ‘xyz’, because they are all multiplied together which makes it one term.

D and E both have 3 terms each, but F has 4 unique terms so this is incorrect also.

G has 3 unique terms, x³, x^4, and 7x, so this is correct.

When H is expanded, you will end up with more than 3 unique terms, so this is incorrect.

I hope this helps!

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X − 3 4 = 8 Which value of x satisfies this equation?

insens350 [35]

. the answer is x=42

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3 years ago

I can't quite remember how to approach this one, I have three others, but I think I could solve them if someone helped walked me

Nimfa-mama [501]

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3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

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3 years ago

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Alex777 [14]

The answer to the question

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