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3 years ago

Expand if necessary combine like terms (7+x)(7-x)=

Mathematics

1 answer:

mina [271]3 years ago

4 0

Answer:

49 - x^2

Step-by-step explanation:

(7+x)(7-x)

We know this is the difference of squares

(a-b) (a+b) = a^2 - b^2

7^2 - x^2 = 49-x^2

We can also FOIL

first = 7*7 = 49

Outer: = 7x

Inner: = -7x

Last : -x*x = -x^2

Add them together

49 +7x-7x - x^2

Combine like terms

49-x^2

You might be interested in

What is the slope of a line passing through (3, 4) and (5,8)?

Licemer1 [7]

Answer:

Step-by-step explanation:

$slope=\frac{y_2-y_1}{x_2-x_1} \\\frac{8-4}{5-3} =\frac{4}{2} =2\\slope=2$

8 0

2 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

3 years ago

If anyone Answer this question with POLITICALLY CORRECT answer I'll give you 25+ Points

Tresset [83]

Answer:

$\sqrt[4]{x^5}$

Step-by-step explanation:

A fraction exponent converts into a radical. The denominator is the index of the radical (farthest left number) and the numerator is the exponent of the base inside (the farthest right number). The base of the fraction exponent is the base number in green. To write this expression, simply the exponents into one exponent and one base.

$x^{\frac{3}{4}}x^{\frac{1}{2} } =x^{\frac{3}{4}}x^{\frac{2}{4} } =x^{\frac{3+2}{4}} =x^{\frac{5}{4}}$