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lutik1710 [3]
3 years ago
9

Find the value of the greater root of x2 - 13x + 12 = 0. A) -12 B) -1 C) 1 D) 12

Mathematics
2 answers:
NISA [10]3 years ago
7 0
B) -1


Hope this helps
Alinara [238K]3 years ago
6 0

Answer:

Option D 12 is the answer.

Step-by-step explanation:

Given is a quadratic equation as

x^2-13x+12=0

To find the roots we can factorise this

Since last term = 12, and middle term -13 we can split middle term as -12 and -1

x^2-12x-x+12 =0\\x(x-12)-1(x-12)=0\\(x-12)(x-1)=0\\x=12: x=1

We know that there are two roots as 12 and 1.

The bigger root is obviously 12.

Option D is the answer

Verify:

Substitute 12 in the equation and check

12^2-13(12)+12=144-156+12=0

Hence our answer is right.

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CAN SOMEONE PLEASE SOLVE THIS????????
WARRIOR [948]

Answer:

27; C; B; D

Step-by-step explanation:

1. B = 120 degrees, and C = 60 degrees. Q = 90 degrees, because B + C = 180 degrees, and the triangle BQC has angle B and angle C bisected. If B = 120 degrees, then the height is equal to 9 sqrt 3, which makes BC equal to 27.

2.C

3. B (You don't actually know if any of the sides are parallel.)

4. D

4 0
3 years ago
How do you do #1 & #2 ?
Alexeev081 [22]
Answer: \\ 1. \: x = \frac{ {32}^{2} }{24} = \frac{128}{3} \\y = \sqrt{ {24}^{2} + {32}^{2} } = 40\\ z = \frac{32 \times 40}{24} = \frac{160}{3} \\ \\ 2. \: x = \sqrt{( {4 \sqrt{5} )}^{2} - {8}^{2} } = 4
7 0
2 years ago
Please help with a, b, and c!!
padilas [110]
(a). 
The product of two binomials is sometimes called FOIL.
It stands for ...

       the product of the First terms                (3j  x  3j)
plus
       the product of the Outside terms          (3j  x  5)
plus
       the product of the Inside terms            (-5  x  3j)
plus
       the product of the Last terms                (-5  x  5)

FOIL works for multiplying ANY two binomials (quantities with 2 terms).

Here's another tool that you can use for this particular problem (a).
It'll also be helpful when you get to part-c .

Notice that the terms are the same in both quantities ... 3j and 5 .
The only difference is they're added in the first one, and subtracted
in the other one.

Whenever you have     

              (the sum of two things) x (the difference of the same things)

the product is going to be

                 (the first thing)²  minus  (the second thing)² .

So in (a), that'll be      (3j)² - (5)²  =  9j² - 25 .

You could find the product with FOIL, or with this easier tool.
______________________________

(b).
This is the square of a binomial ... multiplying it by itself.  So it's
another product of 2 binomials, that both happen to be the same:

                            (4h + 5) x (4h + 5)  .

You can do the product with FOIL, or use another little tool:

The square of a binomial        (4h + 5)²    is ...

         the square of the first term               (4h)²
plus
         the square of the last term                (5)²
plus
         double the product of the terms      2 · (4h · 5)
________________________________

(c).
Use the tool I gave you in part-a . . . twice .

The product of the first 2 binomials is           (g² - 4) .

The product of the last 2 binomials is also    (g² - 4) .

Now you can multiply these with FOIL,
or use the squaring tool I gave you in part-b .

5 0
3 years ago
Read 2 more answers
How do you round 1.038 to the nearest hundredth
Talja [164]
You have to go past the decimal point in since the first number in the tenths place is a zero you can't really round here so you go to the hundreds place and not so three you round down because 3 is closer to 0 then it is 10 so it would be 1. 4



3 0
3 years ago
Read 2 more answers
Help with 21 would be much appreciated.
umka2103 [35]

Answer:

\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

\dfrac{AD}{CD}=\dfrac{CD}{DB}

We have

AD=14-8=6\\CD=y\\DB=8

Substitute:

\dfrac{6}{y}=\dfrac{y}{8}         <em>cross multiply</em>

y^2=(6)(8)

y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}

For x use the Pythagorean theorem:

x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}

3 0
2 years ago
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