All categories

2 years ago

Can someone help me with this question?

Mathematics

1 answer:

kap26 [50]2 years ago

4 0

$\frac{4x + 5 - 5x}{ {6x}^{2} - x - 12 } \\ = \frac{ - x + 5}{ {6x}^{2} - x - 12}$

The answer is the first one.

Hope this helps. - M

You might be interested in

Which of the following best describes the equation below? y=-6x+7

Hitman42 [59]

Answer:

y=-6x+7 (Negative Slope)

Step-by-step explanation:

This equation is in slope intercept form.

7= y-intercept

-6= slope

This means that when you plot this on a graph, your slope will be negative.

5 0

2 years ago

Which undefined term is needed to define a circle?

Solnce55 [7]

A Point is needed to define a circle Hope this helps!!!

3 0

3 years ago

Felicia wants to build a kite with the shape shown. If AC is 60 cm, how many centimeters are in the length of BD?

Kay [80]

Answer:

Step-by-step explanation:

By applying tangent rule in the given right triangle AOB,

tan(30°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

$\frac{1}{\sqrt{3}}=\frac{BO}{OA}$

$OA=BO(\sqrt{3})$

By applying tangent rule in the given right triangle BOC,

tan(60°) = $\frac{OC}{BO}$

OC = BO(√3)

OA + OC = AC

$BO(\sqrt{3})+BO(\sqrt{3}) =60$

2√3(BO) = 60

BO = 10√3

OC = BO(√3)

OC = (10√3)(√3)

OC = 30

By applying tangent rule in right triangle DOC,

tan(60°) = $\frac{OD}{OC}$

OD = OC(√3)

OD = 30√3

Since, BD = BO + OD

BD = 10√3 + 30√3

BD = 40√3

≈ 69.3

6 0

2 years ago

If for two non-zero vectors a,b : ∣a+b∣=∣a−b∣ then angle between them is:______a. 0b. 45c. 60d. 90

Eddi Din [679]

Answer:

$90^\circ$

Step-by-step explanation:

Given two non zero vectors, $\vec a, \vec b$ .

Let the angle between the two vectors = $\theta$

Given that:

$|\vec a+\vec b|=|\vec a-\vec b|$

Let us have a look at the formula for magnitude of addition of two vectors:

$|\vec x+\vec y|=\sqrt{x^2+y^2+2xycos\theta}$

Where $\theta$ is the angle between the two vectors.

formula for magnitude of subtraction of two vectors:

$|\vec x-\vec y|=\sqrt{x^2+y^2-2xycos\theta}$

As per the given condition:

$\sqrt{a^2+b^2+2abcos\theta}=\sqrt{a^2+b^2-2abcos\theta}$

Squaring both sides:

$a^2+b^2+2abcos\theta=a^2+b^2-2abcos\theta\\\Rightarrow 2abcos\theta=-2abcos\theta\\\Rightarrow cos\theta = -cos\theta\\\Rightarrow 2cos\theta = 0\\\Rightarrow cos\theta = 0\\\Rightarrow \theta = 90^\circ$

So, the angle between the two vectors is: $90^\circ$

7 0

2 years ago

Find the value of x for which ABCD must be a parallelogram. 6 54 3 27

stiv31 [10]

Answer:

Step-by-step explanation:

As it's a parallelogram the cross splits each diagonal exactly in half, so each half is the same:

-9-6x = x-30

7x=21

x=3

3 0

2 years ago