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disa [49]
3 years ago
13

Evaluate x^2 − 4x + 5, when x = − 3

Mathematics
2 answers:
tigry1 [53]3 years ago
6 0

Answer:

\boxed{26}

Step-by-step explanation:

\mathsf{ {x}^{2}  - 4x + 5}

\mathrm{Plug \: the \: value \: of \: x}

⇒{( - 3)}^{2}  - 4 \times(- 3 )+ 5

\mathrm{Evaluate \: the \: power}

⇒\mathsf{9 - 4 \times(- 3 ) + 5}

\mathrm{Multiply \: the \: numbers}

⇒\mathsf{9 + 12 + 5}

\mathrm{Add  the  numbers}

⇒\mathsf{26}

Hope I helped!

Best regards!

Gnom [1K]3 years ago
3 0

Answer:

\huge\boxed{26}

Step-by-step explanation:

\sf x^2-4x+5\\Given \ that \ x = -3\\(-3)^2-4(-3)+5\\9+12+5\\26

You might be interested in
Use quadratic formula solve .7x^2 - 28x - 7​
beks73 [17]

Answer:

x = 2 + sqrt(5) or x = 2 - sqrt(5)

Step-by-step explanation using the quadratic formula:

Solve for x over the real numbers:

7 (x^2 - 4 x - 1) = 0

Divide both sides by 7:

x^2 - 4 x - 1 = 0

Add 1 to both sides:

x^2 - 4 x = 1

Add 4 to both sides:

x^2 - 4 x + 4 = 5

Write the left hand side as a square:

(x - 2)^2 = 5

Take the square root of both sides:

x - 2 = sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

x = 2 + sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

Answer:  x = 2 + sqrt(5) or x = 2 - sqrt(5)

6 0
3 years ago
50 POINTS !!<br><br><br> PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.
NeTakaya
I think answer is 9.3 after it’s rounded
use the formula a^2 +b^2=c^2
5 0
2 years ago
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
3 years ago
Please help i really need help <br> thanks
Grace [21]

Answer:

B. 31.62

Step-by-step explanation:

First, find the distance of the line using the distance formula, d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1} )^2   }. Plug the two points (-1,0) and (5,2) into the formula and solve. This gives you 6.32, then this is only one line of a pentagon, which has 5 sides, so you need to multiply. Since you are finding the perimeter, which is all of the sides added together, multiply the one side by 5. 6.32*5=31.62.

6 0
3 years ago
5
lana66690 [7]

Answer:

a

Step-by-step explanation:

i did  this in middle shoool

hehe

6 0
2 years ago
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