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3 years ago

How could Brent use a rectangle to model the factors of x2 – 7x + 6?

2 answers:

Crank3 years ago

6 0

He could draw a diagram of a rectangle with dimensions x – 1 and x – 6 and then show the area is equivalent to the sum of x2, –x, –6x, and 6.

ivann1987 [24]3 years ago

6 0

Answer:

Step-by-step explanation:

Since Brent uses a rectangle that models the area = (x² - 7x + 6)

Now we will factorize this expression

x² - 7x + 6 = x² - 6x - x + 6

= x(x - 6) - 1(x - 6)

= (x - 1)(x - 6)

Therefore, (x -1) and (x - 6) will represent the sides of a rectangle so that multiplication of the sides will model the area of the rectangle.

You might be interested in

If h(x)=4x^2-16 we’re shifted 7 units to the right and 3 units down what would the new equation be

attashe74 [19]

Answer:

4(x-7)^2 - 19

Step-by-step explanation:

Shifting 7 units to the right: 4(x-7)^2 - 16.

3 units down: 4(x-7)^2 - 16 - 3, or 4(x-7)^2 - 19

6 0

3 years ago

The total length of wire needed to fasten 175 bundles is 2,975 cm.

UkoKoshka [18]

According to the conversion table we know,

1 meter = 100cm

This means if we need to convert meters to centimeters, we need to multiply by 100

If we need to convert from centimeters to meters, we need to divide by 100.

Here the length of the wire needed to wrap 175 bundles = 2975 cm.

Now as we need to convert it to meters.

We need to divide this by 100.

We get

2975/100= 29.75

So 2975 cm = 29.75 m

Option D) is the right answer

3 0

3 years ago

Read 2 more answers

A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square roo

deff fn [24]

Answer:

$\frac{x^2}{61}-\frac{y^2}{37} =1$

Step-by-step explanation:

The standard equation of a hyperbola is given by:

$\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1$

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

$\frac{x^2}{61}-\frac{y^2}{37} =1$

6 0

3 years ago

Can someone pls help me

Sophie [7]

Answer:

Step-by-step explanation:

If you divide 36 by 12, you get 3.

For this to work, whatever the white wax is, it needs to be 3 times as more as 5.

5x3=15

5 0

3 years ago

Read 2 more answers

Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

5 0

3 years ago