All categories

4 years ago

Evaluate 3X^2-14X-49/X-7 the quotient of the equation is X+

Mathematics

2 answers:

jenyasd209 [6]4 years ago

6 0

Answer:

$3x+7$

Step-by-step explanation:

$\frac{3x^2-14x-49}{x-7}$

before dividing this fraction , lets factor the numerator

$3x^2-14x-49$

Product is -49 and sum is -14, lets break the middle term using factors

$(3x^2+7x)+ (-21x-49)$

Take out GCF from each group

$x(3x+7)-7(3x+7)$

$(3x+7)(x-7)$

Now we replace the factors in the numerator

$\frac{(3x+7)(x-7)}{x-7}$

We have (x-7) at the top and bottom , so we cancel it out

$3x+7$

denpristay [2]4 years ago

5 0

I hope this helps you

You might be interested in

What is cos(45°)?

Andrej [43]

Answer:

$\rm \dfrac{\sqrt2}{2}$

Step-by-step explanation:

A right angled ∆ is given to us and we need to find out the value of cos 45° ( with rationalized denominator ) . As we know that in a right angle ∆ , the value of cosine is ,

$\rm\implies cos\theta =\dfrac{base}{hypotenuse}$

In the given triangle ,

Value of base = 1
Value of hypotenuse = √2

On substituting the respective values ,

$\rm\implies cos45^\circ =\dfrac{1}{\sqrt2}$

For rationalizing the denominator , multiply numerator and denominator by √2 , we have,

$\rm\implies cos45^\circ=\dfrac{1(\sqrt2)}{(\sqrt2)(\sqrt2)}$

Simplify ,

$\rm\implies \boxed{\pink{\frak{ cos45^\circ=\dfrac{\sqrt2}{2} }}}$

6 0

3 years ago

Find the exact value of cos(a+b) if cos a=-1/3 and cos b=-1/4 if the terminal side if a lies in quadrant 3 and the terminal side

maria [59]

Answer:

cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = $-\frac{1}{3}$

cos(b) = $-\frac{1}{4}$

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = $-\sqrt{1-(-\frac{1}{3})^2}$ [Since, sin(a) = $\sqrt{(1-\text{cos}^2a)}$ ]

= $-\sqrt{\frac{8}{9}}$

= $-\frac{2\sqrt{2}}{3}$

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be negative.

sin(b) = $-\sqrt{1-(-\frac{1}{4})^2}$

= $-\sqrt{\frac{15}{16}}$

= $-\frac{\sqrt{15}}{4}$

By substituting these values in the identity,

cos(a + b) = $(-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})$

= $\frac{1}{12}-\frac{\sqrt{120}}{12}$

= $\frac{1}{12}(1-\sqrt{120})$

= $\frac{1}{12}(1-2\sqrt{30})$

Therefore, cos(a + b) = $\frac{1}{12}(1-2\sqrt{30})$

5 0

3 years ago

triangle ABC is Isosceles CA and CB are equal AB is parallels to the x- axis. Work out the coordinate of A. Given that BC=5cm wo

BabaBlast [244]

Answer:

The perimeter of the triangle ABC is 17 cm.

Step-by-step explanation:

Consider the Isosceles triangle ABC.

The sides CA and CB are equal with measures, 5 cm.

The base angles are assumed to be x° each. Hence, the angle ACB is 2x°.

The altitude CP divides the base AB into two equal halves and the angle ACB is also cut into halves.

Consider the right angled triangle ACP.

The sum of all the angles in a triangle is 180°.

Determine the value of x as follows:

x° + x° + 90° = 180°

2x° = 90°

x° = 45°

Compute the length of side AP as follows:

$cos\ 45^{0}=\frac{AP}{CA}$

$\frac{1}{\sqrt{2}}=\frac{AP}{5}$

$AP =\frac{5}{\sqrt{2}}\\\\AP=3.5$

Then the length of side AB is:

AB = AP + PB

= 3.5 + 3.5

= 7 cm

The perimeter of triangle ABC is:

Perimeter = AB + CA + CB

= 7 + 5 + 5

= 17

Thus, the perimeter of the triangle ABC is 17 cm.

8 0

3 years ago

To make 3/4 pound of mixed nuts, how many pounds of cashews would you add to 1/8 pound of almonds and 1/4 pound of peanuts?

Anna11 [10]

you will need 3/8 pounds of peanut.

3/8 + 1/4 + 1/8= 3/4

(ask me if you are confuse plz)

7 0

3 years ago

You decide to deposit $5,000 in a savings account that earns 3% interest. How much money will you have in 10 years

ki77a [65]

Answer:

65,000

Step-by-step explanation:

7 0

3 years ago

Evaluate 3X^2-14X-49/X-7 the quotient of the equation is __X+__

Evaluate 3X^2-14X-49/X-7 the quotient of the equation is X+