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3 years ago

Solve for x. x2 − 9x + 20 = 0

Mathematics

2 answers:

nordsb [41]3 years ago

7 0

Answer:

x=4 or x=5

Step-by-step explanation:

Let's solve your equation step-by-step.

x2−9x+20=0

Step 1: Factor left side of equation.

(x−4)(x−5)=0

Step 2: Set factors equal to 0.

x−4=0 or x−5=0

x=4 or x=5

umka2103 [35]3 years ago

3 0

x has two values 4 and 5 by factorization or using the general formula

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The measures of complementary angles have a sum of 90 degrees.Angle A and angle B are complimentary, and their measures have a d

777dan777 [17]

Answer:

The measure of the angles are 55° and 35°

Step-by-step explanation:

Sum of two angles that are complement of each other are 90.

The angles are A and B, so we can write:

A + B = 90

Also,

The difference is 20. Let A be the greater of them, so we can write equation 2 as:

A - B = 20

Now, we add up both equations, eliminate B and solve for A first:

A + B = 90

A - B = 20

---------------

2A = 90 + 20

2A = 110

A = 110/2

A = 55

And now the measure of B:

A + B = 90

55 + B = 90

B = 90 - 55

B = 35

5 0

4 years ago

Give a vector parametric equation for the line through the point (−2,−4,0) that is parallel to the line ⟨−2−3t,1−4t,−5t⟩:

eimsori [14]

The tangent vector for the given line is

T(t) = d/dt ⟨-2 - 3t, 1 - 4t, -5t⟩ = ⟨-3, -4, -5⟩

On its own, this vector points to a single point in space, (-3, -4, -5).

Multiply this vector by some scalar t to get a whole set of vectors, essentially stretching or contracting the vector ⟨-3, -4, -5⟩. This set is a line through the origin.

Now translate this set of vectors by adding to it the vector ⟨-2, -4, 0⟩, which correspond to the given point.

Then the equation for this new line is simply

L(t) = ⟨-3, -4, -5⟩t + ⟨-2, -4, 0⟩ = ⟨-2 - 3t, -4 - 4t, -5t⟩

3 0

3 years ago

During a research experiment, it was found that the number of bacteria in a culture grew at a rate proportional to its size. At

BartSMP [9]

No of bacteria present at midnight. = 6700+1680=8380

7 0

4 years ago

Ayudaaaaa porfavor Convertir las unidades de medida 8 000 m = _______ km 4 000 g = _______ kg 3 000 m = ________ km 7 m = ______

jek_recluse [69]

Answer:

a) $x = 8\,km$ , b) $x = 4\,kg$ , c) $x = 3\,km$ , d) $x = 700\,cm$ , e) $x = 3\,cm$ , f) $x = 100\,mm$ , g) $x = 7\,mm$ , h) $x = 10000\,mL$ , i) $x = 8\,m$ , j) $x = 6000\,g$

Step-by-step explanation:

a) 1 kilómetro equivale a 1000 metros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 8000\,m \times \frac{1\,km}{1000\,m}$

$x = 8\,km$

b) 1 kilogramo equivale a 1000 gramos, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 4000\,g \times \frac{1\,kg}{1000\,g}$

$x = 4\,kg$

c) 1 kilómetro equivale a 1000 metros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 3000\,m \times \frac{1\,km}{1000\,m}$

$x = 3\,km$

d) 1 metro equivale a 100 centímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 7\,m \times \frac{100\,cm}{1\,m}$

$x = 700\,cm$

e) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 30\,mm \times \frac{1\,cm}{10\,mm}$

$x = 3\,cm$

f) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 10\,cm \times \frac{10\,mm}{1\,cm}$

$x = 100\,mm$

g) 1 centímetro equivale a 10 milímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 70\,mm\times \frac{1\,cm}{10\,mm}$

$x = 7\,mm$

h) 1 litro equivale a 1000 mililitros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 10\,L\times \frac{1000\,mL}{1\,L}$

$x = 10000\,mL$

i) 1 metro equivale a 100 centímetros, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 800\,cm \times \frac{1\,m}{100\,cm}$

$x = 8\,m$

j) 1 kilogramo equivale a 1000 gramos, entonces obtenemos el valor requerido mediante la siguiente regla de tres simple:

$x = 6\,kg\times \frac{1000\,g}{1\,kg}$

$x = 6000\,g$

8 0

3 years ago

I really stuck on trying to prove this

Alisiya [41]

$\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)$

When $n=1$ ,

$\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!$

Meanwhile, you have on the right

$\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p!$

so the equality holds for $n=1$ .

Assume it holds for $n=k$ , i.e. that

$\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}$

Now for $n=k+1$ , you have

$\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}$

as required.

3 0

3 years ago