How do you solve...0=-5t^2+20t+1

Answers

Answer 1

Answer: there were 2 solutions that i came up with. Here is the first one. I rearranged the eqaution by subtracting what is to the right of thr equal sign. Multiply the coefficeint of the first term by the constant 5 x (-1)= -5. Then you would find 2 factors of -5 whose sum equals the coefficient of the middle term which is 20. -5+1= -4 and -1+5=4. the eqaution then comes to 5t^2-20t-1=0. you would solve 5t^2-20t-1=0. you would divide both sides of the equal sign by 5. t^-4t-1/5=0. then t^2-4t=1/5. add 4 to both sides of the equation. so we get 21/5+ 4 + 4t -t^2=21/5. it then comes out to be t^2-4t+t=t-2^2. according to the law of transitivity it is t-2^2=21/5.

t =(20+√420)/10=2+1/5√ 105 = 4.049 .

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What are the solutions to the equation x-7/x=6? a. x=-7 and x=1
b. x=-6 and x=-1
c. x=-1 and x=7
d. x=1 and x=6

Answers

$x - (7)/(x) = 6$
$(x^(2))/(x) - (7)/(x) = 6$
$(x^(2) - 7)/(x) = 6$
$x^(2) - 7 = 6x$
$x^(2) - 6x - 7 = 0$
$x^(2) - 7x + x - 7 = 0$
$x(x) - x(7) + 1(x) - 1(7) = 0$
$x(x - 7) + 1(x - 7) = 0$
$(x + 1)(x - 7) = 0$
$x + 1 = 0$ $or$ $x - 7 = 0$
$x = -1$ $or$ $x = 7$

Answer:

Step-by-step explanation:

The equation is $x-(7)/(x)=6$

LCM of the denominator on the left side gives us:

$(x^(2)-7)/(x)=6$

Cross multiplying and arranging gives us:

$x^(2)-7=6x\nx^(2)-6x-7=0$

This is a binomial, to factor this, we have to think of 2 numbers that multiplied gives us $-7$ [the constant term] and added gives us $-6$ [the coefficient of x].

The two numbers are: $-7$ and $1$ .

Now we can write:

$x^(2)-6x-7=0\n(x-7)(x+1)=0$

Hence, either $(x-7)=0$ OR $(x+1)=0$ . This gives us two solutions, $x=7$ and $x=-1$ . Answer choice C is correct.

Solve the system of equations using cramer's rule -x+y-3z=-4 3x-2y+8z=14 2x-2y+5z=7

Answers

System of Equations
$-1x + 1y - 3z = -4 \n3x - 2y + 8z = 14 \n2x - 2y + 5z = 7$

Coefficient Matrix's Determinant
$D = \left[\begin{array}{ccc}-1&1&-3\n3&-2&8\n2&-2&5\end{array}\right]$

Answer Column
$\left[\begin{array}{ccc}-4\n14\n7\end{array}\right]$

Dx: Coefficient Determinant with Answer-Column values in X-Column
$D_(x) = \left[\begin{array}{ccc}-4&1&-3\n14&-2&8\n7&-2&5\end{array}\right]$

Dy: Coefficient Determinant with Answer-Column Values in Y-Column
$D_(y) = \left[\begin{array}{ccc}-1&-4&-3\n3&14&8\n2&7&5\end{array}\right]$

Dz: Coefficient Determinant with Answer-Column Values in Z-Column
$D_(z) = \left[\begin{array}{ccc}-1&1&-4\n3&-2&14\n2&-2&7\end{array}\right]$

Evaluating each Determinant
$D= \left[\begin{array}{ccc}-1&1&-3\n3&-2&8\n2&-2&5\end{array}\right] \nD = (-1 * (-2) * 5) + (1 * 8 * 2) + (-3 * 3 * (-2)) - (2 * (-2) * (-3)) - (-2 * 8 * (-1)) - (5 * 3 * 1) \nD = (10) + (16) + (18) - (12) - (16) - (15) \nD = 10 + 16 + 18 - 12 - 16 - 15 \nD = 26 + 18 - 12 - 16 - 15 \nD = 44 - 12 - 16 - 15 \nD = 32 - 16 - 15 \nD = 16 - 15 \nD = 1$

$D_(x) = \left[\begin{array}{ccc}-4&1&-3\n14&-2&8\n7&-2&5\end{array}\right] \nD_(x) = (-4 * (-2) * 5) + (1 * 8 * 7) + (-3 * 14 * (-2)) - (7 * (-2) * (-3)) - (-2 * 8 * (-4)) - (5 * 14 * 1)) \nD_(x) = (40) + (56) + (84) - (42) - (64) - (70) \nD_(x) = 40 + 56 + 84 - 42 - 64 - 70 \nD_(x) = 96 + 84 - 42 - 64 - 70 \nD_(x) = 180 - 42 - 64 - 70 \nD_(x) = 138 - 64 - 70 \nD_(x) = 74 - 70 \nD_(x) = 4$

$D_(y) = \left[\begin{array}{ccc}-1&-4&-3\n3&14&8\n2&7&5\end{array}\right] \nD_(y) = (-1 * 14 * 5) + (-4 * 8 * 2) + (-3 * 3 * 7) - (2 * 14 * (-3)) - (7 * 8 * (-1)) * (5 * 3 * (-4)) \nD_(y) = (-70)+ (-64) + (-63) - (-84) - (-56) - (-60) \nD_(y) = -70 - 64 - 63 + 84 + 56 + 60 \nD_(y) = -134 - 63 + 84 + 56 + 60 \nD_(y) = -197 + 84 + 56 + 60 \nD_(y) = -113 + 56 + 60 \nD_(y) = -57 + 60 \nD_(y) = 3$

$D_(z) = \left[\begin{array}{ccc}-1&1&-4\n3&-2&14\n2&-2&7\end{array}\right] \nD_(z) = (-1 * (-2) * 7) + (1 * 14 * 2) + (-4 * 3 * (-2)) - (2 * (-2) * (-4)) - (-2 * 14 * (-1)) - (7 * 3 * 1) \nD_(z) = (14) + (28) + (24) - (16) - (28) - (21) \nD_(z) = 14 + 28 + 24 - 16 - 28 - 24 \nD_(z) = 42 + 24 - 16 - 28 - 21 \nD_(z) = 66 - 16 - 28 - 21 \nD_(z) = 50 - 28 - 21 \nD_(z) = 22 - 21 \nD_(z) = 1$

$x = (D_(x))/(D) = (4)/(1) = 4 \ny = (D_(y))/(D) = (3)/(1) = 3 \nz = (D_(x))/(D) = (1)/(1) = 1 \n(x, y, z) = (4, 3, 1)$

What is the explicit formula

Answers

Answer:explicit formula is a formula we can use to find the nth term of a sequence. In the easiest definition, explicit means exact or definite. The formula is explicit because as long as it's applied correctly, the nth term can be determined

Step-by-step explanation:

Answer:

inkly dinkly

Step-by-step explanation:

moopsie poopsie

When you use the distance formula, does the order in which you subtract the x- and y- coordinates matter? Explain please

Answers

lets say ur points are (1,2)(3,4)

d = sqrt (x2 - x1)^2 + (y2 - y1)^2

u would label ur points
set 1 (1,2)...x1 = 1 and y1 = 2
set 2 (3,4)...x2 = 3 and y2 = 4

or u could label them this way..
set 1 (3,4)...x1 = 3 and y1 = 4
set 2 (1,2)...x2 = 1 and y2 = 2

u have to do (x2 - x1)^2 and (y2- y1)^2....but u can label ur sets of points different. Just make sure u keep ur sets together. Does that make sense ?

The time it takes for an object stopped from a certain speed can be modeled by the equationt=1/2 square root v, where v is the speed of object in meters per second. if it takes 3 seconds for the object to stop, what is the speed of object in meters per second?

Answers

Given:
t = 1/2 √v
v = speed of object meters per second

t = 3 seconds

3 = 1/2 √v
3 * 2 = √v
6 = √v
6² = √v²
36 meters = v

v = 36 meters per second

What's the answer for this question.

Answers

You're looking for a point that is not in the region where both functions touch (darker shade).

Points A, B, and D all fall into that region (even though you can't see it in the graph for D).

Your answer is C.

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