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3 years ago

What's the answer to x^2-6x-8?

Mathematics

1 answer:

solong [7]3 years ago

6 0

Answer:

Step-by-step explanation:

(x-4)(x-2)=0

If any individual factor on the left side of the equation is equal to 0

the entire expression will be equal to 0

Set the first factor equal to 0 and solve

x=4

The final solution is all the values that make

(x-4)(x-2)=0

x=4,2

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HELP ME PLEASE!

Delicious77 [7]

Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.

Let's see whether they do.

8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.

If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:

8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and

d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.

11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.

7 0

3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

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3 years ago

Read 2 more answers

Calculate the next 3 terms and write the formula for the nth term for the following sequences. x, x+2, x+4,…

Vikki [24]

Answer:

Forth term = x + 6

Fifth term = x + 8

Sixth term = x + 10

Formular for nth term:

${ \tt{n _{th} = a + (n - 1)d}} \\ { \tt{{n _{th} = x + (n - 1) \times 2}}} \\ { \tt{{n _{th} =x + 2n - 2 }}}$

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3 years ago

Read 2 more answers

Jo borrowed 3,800 for 4 months from a bank. the bank discounted the loan at 5.5%.

garri49 [273]

A.
I=prt
I=(3800)(.055)(4/12)
I=$69.67

b.
Principal-interest=money received from Bank
3800-69.67=$3730.34

c.
I=prt
69.67=3730.34(r)(4/12)
69.67=1243.45r
69.67/1243.45=1243.45/1243.45r
.056≈r
Rate is 5.60%

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3 years ago

Which number sentence and model shows the quotient of 2/5 divided by 2/3 select two correct answers

murzikaleks [220]

Answer:

I can't understand so don't worry

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2 years ago